lecture 07 w, e and sm's (1)

WorkEnergySimple

Machines

A 50 kg box slides across a floor and its velocity decreases from 20 m/s to 5 m/s. a) What change in Kinetic Energy (ΔKE) did the box undergo to reduce its velocity? ΔKE = ½ (m)(vf

2 – vi2)

ΔKE = ½ (50 kg)[(5 m/s)2 – (20 m/s)2]

ΔKE = ½ (50 kg)[-375 m2/s2]

ΔKE = -9375 kgm2/s2

ΔKE = -9375 J

A 50 kg box slides across a floor and its velocity decreases from 20 m/s to 5 m/s. b) How much work does the frictional resistance of the floor do upon the block?

Work Done by Frictional Forces Is EQUAL and OPPOSITE

The Change in Kinetic Energy

+9375 J

c) If the reduction in speed takes place over 20 m, what is the force of friction?

Work (J) = Force (N) X distance (m)

Force (N)= [Work (J) / distance

(m)]

NNForce

m

JNForce

mcedis

JWorkNForce

469)(

20

9375)(

)(tan

)()(

Exercise 1:On his way off to college, Russell drags his suitcase 15.0 m from the door of his house to the car at a constant speed with a horizontal force of 95.0 N.a) How much work does Russell do to

overcome the force of friction?b) If the floor has just been waxed, does he have to do more work or less work to move the suitcase? Explain.

𝒂¿𝑾=𝑭 ∆𝒅=(𝟗𝟓 .𝟎𝑵 ) (𝟏𝟓 .𝟎𝒎 )=𝟏𝟒𝟑𝟎 𝑱

Exercise 2: Katie, a 30.0-kg child, climbs a tree to rescue her cat who is afraid to jump 8.0 m to the ground. How much work does Katie do in order to reach the cat?

Exercise 3: Marissa does 3.2 J of work to lower the window shade in her bedroom adistance of 0.8 m. How much force must Marissa exert on the window shade?

Exercise 4: Atlas and Hercules, two carnival sideshow strong men, each lift 200. kg barbells 2.00 m off the ground. Atlas lifts his barbells in 1.00 s and Hercules lifts his in 3.00 s.

a) Which strong man does more work? man is more powerful.

Hercules and Atlas…Same Work…

Exercise 4: Atlas and Hercules, two carnival sideshow strong men, each lift 200. kg barbells 2.00 m off the ground. Atlas lifts his barbells in 1.00 s and Hercules lifts his in 3.00 s.

b) Calculate which man is more powerful.

Hercules

Atlas

Example 7: Frank, a San Francisco hot dog vender, has fallen asleep on the job. When an earthquake strikes, his 300 kg hot dog cart rolls down Nob Hill and reaches point A at a speed of 8.00 m/s. How fast is the hot-dog cart going at point B when Frank finally wakes up and starts to run after it?

𝑲𝑬𝟎+𝑷𝑬𝟎=𝑲𝑬𝒇 +𝑷𝑬𝒇

(𝟏𝟐 )𝒎𝒗𝟎𝟐+𝒎𝒈𝒉𝟎=(𝟏𝟐 )𝒎𝒗 𝒇 𝟐+𝒎𝒈𝒉 𝒇

Mass is same throughout and can be factored out…

(𝟏𝟐 )𝒗 𝟎𝟐+𝒈𝒉𝟎=(𝟏𝟐 )𝒗 𝒇 𝟐+𝒈𝒉 𝒇

Solve for vf

=

=

v0 = 8 m/sg = 10 m/s2

Unknown: vf = 8 m/sOriginal Equation ΔKE = ΔPE

=

h0 = 50 mhf = 30 m

=

=

=

Exercise 5:It is said that Galileo dropped objects off the Leaning Tower of Pisa to determine whether heavy or light objects fall faster. If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, what would have been the change in PE of the cannon ball?

∆ 𝑷𝑬=𝒎𝒈𝒉∆ 𝑷𝑬=(𝟓𝒌𝒈)(𝟏𝟎𝒎

𝒔𝟐 )(𝟏𝟐𝒎)

∆ 𝑷𝑬=𝟔𝟎𝟎 𝑱

Exercise 6: On June 5, 2007, Rags to Riches, became the first filly to win the Belmont Stakes horse race since 1905, running with an average speed of 16.23 m/s. If Rags to Riches and jockey Johnny Velazquez had a combined mass of 550.0 kg, what was their KE as they crossed the finish line?

∆ 𝑲𝑬=𝟏𝟐

𝒎𝒗𝟐

∆ 𝑲𝑬=𝟏𝟐

(𝟓𝟓𝟎𝒌𝒈)(𝟏𝟔 .𝟐𝟑𝒎𝒔 )

𝟐

∆ 𝑲𝑬=𝟕𝟐 ,𝟒𝟒𝟎 𝑱

Exercise 7: Brittany is changing the tire of her car on a steep hill 20.0 m high. She trips and drops the 10.0-kg spare tire, which rolls down the hill with an initial speed of 2.00 m/s. What is the speed of the tire at the top of the next hill, which is 5.00 m high?

(𝟏𝟐 )𝒎𝒗𝟎𝟐+𝒎𝒈𝒉𝟎=(𝟏𝟐 )𝒎𝒗 𝒇 𝟐+𝒎𝒈𝒉 𝒇

=

=

=

Exercise 8: A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. a) If a bean of mass 2.0 g jumps 1.0 cm from your hand into the air, how much potential energy has it gained in reaching its highest point. b) What is its speed as the bean lands back in the palm of your hand?

mv2

=

∆ 𝑷𝑬=(𝟎 .𝟎𝟎𝟐𝐤𝐠)(𝟏𝟎 𝒎𝒔𝟐 )(𝟎 .𝟎𝟏𝐦)

J =

𝒗=√¿ ¿

Exercise 9: A 500.-kg pig is standing at the top of a muddy hill on a rainy day. The hill ha a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. What is the pig’s speed at the bottom of the hill? Use the law of conservation of energy.

𝒗=𝟐𝟒 .𝟓𝒎/𝒔

∆ 𝑲𝑬=∆𝑷𝑬

𝒗=√𝟐𝒈𝒉 𝒗=√𝟐 (𝟏𝟎𝒎𝒔𝟐 )(𝟑𝟎𝒎)

𝟏𝟐

𝒗𝟐=𝒈𝒉𝟏𝟐

𝒎𝒗𝟐=𝒎𝒈𝒉

Exercise 10: While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on Earth. If Neil Armstrong jumped up on the moon with an initial speed of 1.51 m/s to a height of 0.700 m, what amount of gravitational acceleration did he experience?

𝒈=(𝟏 .𝟓𝟏𝒎 /𝒔 )𝟐

𝟐 (𝟎 .𝟕𝟎𝟎𝒎)

∆ 𝑲𝑬=∆𝑷𝑬𝟏𝟐

𝒗𝟐=𝒈𝒉𝟏𝟐

𝒎𝒗𝟐=𝒎𝒈𝒉 𝒗𝟐

𝟐𝒈=𝒉

The Sweet SpotThe sweet spot of a softball bat or a tennis racquet is the place where the ball’s impact produces minimum vibrations in the racquet or bat. Strike a ball at the sweet spot and it goes faster and farther. Strike a ball in another part of the bat or racquet, and vibrations can occur that sting your hand! From an energy point of view, there is energy in the vibrations of the bat or racquet. There is energy in the ball after being struck. Energy that is not in vibrations is energy available to the ball. Do you see why a ball will go faster and farther when struck at the sweet spot?

Due to friction, energy is transferred both into the floor and into the tire when the bicycle skids to a stop.

a. An infrared camera reveals the heated tire track on the floor.

b. The warmth of the tire is also revealed.

Typical stopping distances for cars equipped with antilock brakes traveling at various speeds. The work done to stop the car is friction force distance of slide. (Miles per Hour)

28

56

113

(Slide is in meters)10

401 6 0

When released, potential energy will become the kinetic energy of the arrow.

What process provides energy for rockets that lift the space shuttle into orbit? What process releases energy from the food we eat? The answer is chemical reactions.

During a chemical reaction the bonds between atoms break and then reform. Breaking bonds requires energy, and forming bonds releases it. Pulling atoms apart is like pulling apart two magnets stuck together; it takes energy to do it. And when atoms join, it is like two separated magnets that slam together; energy is released.

Rapid energy re lease can produce flames. Slow energy re lease occurs during the digestion of food. The conservation of energy rules chemical reactions. The amount of energy required to break a chemical bond is the same amount released when that bond is formed.

The potential energy of the 100-N boulder with respect to the ground below is 200 J in each case because the work done in elevating it 2 m is the same whether the boulder is…a. Lifted with

100 N of force…

No work is done in moving it horizontally, neglecting friction.

Potential Energy

KE (J) PE (J) ΣFw

(N)d

(m)

500200

010,00

010,00

0

500150

2,500 7,50010,00

0

500100

5,000 5,00010,00

0

500 50 7,500 2,50010,00

0

500 0 10,000 010,00

0

Everywhere along the path of the pendulum bob, the sum of

PE and KE is the same. Because of the work done

against friction, this energy will eventually be transformed

into heat.

AMA =Foutp

ut

Finput

IMA =dinput

doutput

Eff =w

output

winput

Ideal: Winput = Woutput

Real: Winput >

Woutput

b. Pushed up the 4-m incline with 50 N of force.


C. Lifted with 100 N of force up each 0.5-m stair.


The Hard Way

The Machine

=

The Hard Way The Machine Way =

F d F d

=

The work is the

same….with or without a

machine

a. A pulley can change the direction of a force.

b. A pulley multiplies force.

b. A pulley multiplies force.

c. Two pulleys can change the direction and multiply force

A compl

ex pulley syste

m.

Energy Conservation Most energy consumed in America comes from fossil fuels. Oil, natural gas, and coal supply the energy for almost all our industry and technology. About 70% of electrical power in the United States comes from fossil fuels, with about 21% from nuclear power. Worldwide, fossil fuels also account for most energy consumption. We have grown to depend on fossil fuels because they have been plentiful and inexpensive. Until recently, our consumption was small enough that we could ignore their environmental impact.

But things have changed. Fossil fuels are being consumed at a rate that threatens to deplete the entire world supply. Locally and globally, our fossil fuel consumption is measurably polluting the air we breathe and the water we drink. Yet, despite these problems, many people consider fossil fuels to be as inexhaustible as the sun’s glow and as acceptable as Mom’s apple pie, because these fuels lasted and nurtured us through the 1900s.

Financially, fossil fuels are still a bargain, but this is destined to change. Environmentally, the costs are already dramatic. Some other fuel must take the place of fossil fuels if we are to maintain the industry and technology to which we are accustomed. The French have chosen nuclear, with about 74% of their electricity coming from nuclear power plants. What energy source would you choose as an alternative?

In the meantime, we shouldn’t waste energy. As individuals, we should limit the consumption of useful energy by such measures as turning off unused electrical appliances, using less hot water, going easy on heating and air conditioning, and driving energy-efficient automobiles. By doing these things, we are conserving useful energy.

In how many reasonable ways can we reduce energy consumption?

When electric current passes through water, bubbles of hydrogen form at one wire and bubbles of oxygen form at the other.

In a fuel cell, the reverse process occurs: hydrogen and oxygen combine to produce water and electricity.

.Wheel and Axle

Pulley

InclinePlane

Exercise 11: Cathy, a 460-N actress playing Peter Pan, is hoisted above the stage in order to “fly” by a stagehand pulling with a force of 60. N on a rope wrapped around a pulley system. What is the actual mechanical advantage of the pulley system?

𝑨𝑴𝑨=𝑭 𝒐

𝑭 𝒊

𝑨𝑴𝑨=𝟒𝟔𝟎𝑵𝟔𝟎𝑵

𝑨𝑴𝑨=𝟕 .𝟕

Exercise 12: A windmill uses sails blown by the wind to turn an axle that allows a grindstone to grind corn into meal with a force of 90. N. The windmill has sails of radius 6.0 m blown by a wind that exerts a force of 15 N on the sails, and the axle of the grindstone has a radius of 0.50 m. a) What is the ideal mechanical advantage of the wheel?

𝑰𝑴𝑨=𝒅𝒊

𝒅𝒐

𝑰𝑴𝑨=𝟐𝝅 𝒓 𝒔

𝟐𝝅 𝒓𝒂

𝑰𝑴𝑨=𝒓 𝒔

𝒓 𝒂𝑰𝑴𝑨=

𝟔 .𝟎𝒎.𝟓𝒎

𝑰𝑴𝑨=𝟏𝟐

Exercise 12: A windmill uses sails blown by the wind to turn an axle that allows a grindstone to grind corn into meal with a force of 90. N. The windmill has sails of radius 6.0 m blown by a wind that exerts a force of 15 N on the sails, and the axle of the grindstone has a radius of 0.50 m. b) What is the actual mechanical advantage of the wheel?


𝑭 𝒊𝑨𝑴𝑨=

𝟗𝟎𝑵𝟏𝟓𝑵

𝑨𝑴𝑨=𝟔

Exercise 12: A windmill uses sails blown by the wind to turn an axle that allows a grindstone to grind corn into meal with a force of 90. N. The windmill has sails of radius 6.0 m blown by a wind that exerts a force of 15 N on the sails, and the axle of the grindstone has a radius of 0.50 m. c) What is the efficiency of the wheel?

𝑬 𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲=𝑨𝑴𝑨𝑰𝑴𝑨

𝑬 𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲=𝟔𝟏𝟐

𝑬 𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲=𝟎 .𝟓

Exercise 13: Winnie, a waitress, holds in one hand a 5.0-N tray stacked with twelve 3.5-N dishes. The length of her arm from her hand to her elbow is 30.0 cm and her biceps muscle exerts a force 5.0 cm from her elbow, which acts as a fulcrum. How much force must her biceps exert to allow her to hold the tray?Assuming an efficiency of

1…

=𝒅𝒊

𝒅𝒐=

𝑭𝒐

𝑭 𝒊

=

=

Exercise 14: When building the pyramids, the ancient Egyptians were able to raise large stones to very great heights by using inclines. If an incline has an ideal mechanical advantage of 4.00 and the pyramid is 15.0 m tall, how much of an angle would the incline need in order for the Egyptian builder to reach the top? 𝒅𝒊=𝒅𝒐 𝑰𝑴𝑨

𝒅𝒊=(𝟏𝟓𝒎)(𝟒)𝒅𝒊=𝟔𝟎𝒎

𝒅𝒊=𝟔𝟎𝒎

𝒅𝒐=𝟏𝟓𝒎

𝜽=𝒔𝒊𝒏−𝟏 𝟏𝟓𝒎𝟔𝟎𝒎

𝜽=𝟏𝟒 .𝟓𝟎

𝛉

Exercise 15: The Ramseys are moving to a new town, so they have called in the ACME moving company to take care of their furniture. Debbie, one of the movers, slides the Ramseys’ 2200-N china cabinet up a 6.0-m-long ramp to the moving van, which stands 1.0 m off the ground.

a) What is the ideal mechanical advantage of the incline?

IMA

𝒅𝒐=𝟏𝒎

𝛉

IMA

IMA𝒅

𝒐=𝟔𝒎

b) If Debbie must exert a 500.-N force to move the china cabinet up the ramp with a constant speed, what is the actual mechanical advantage of the ramp?

A-1: On a ski weekend in Colorado, Bob, whose mass is 75.0 kg, skis down a hill that is inclined at an angle of 15.0° to the horizontal and has a vertical rise of 25.0 m. How much work is done by gravity on Bob as he goes down the hill?

𝒘=𝑭𝒅

𝒘=𝟏𝟖𝟖𝟎𝟎 𝑱

𝒘=𝒎𝒈𝒅𝒘=𝟕𝟓𝒌𝒈 (𝟏𝟎𝒎

𝒔𝟐 )𝟐𝟓 .𝟎𝒎

25.0 m.

c) What is the efficiency of the ramp?

𝑬𝒇𝒇 =𝑨𝑴𝑨𝑰𝑴𝑨

𝑬𝒇𝒇 =𝟒 .𝟒𝟔

𝑬𝒇𝒇 =.𝟕𝟑

A-2: A pile driver is a device used to drive stakes into the ground. While building a fence, Adam drops a pile driver of mass 3000. kg through a vertical distance of 8.0 m. The pile driver is opposed by a resisting force of 5.0 x 106 N. How far is the stake driven into the ground on the first stroke?

𝑭 𝒐𝒑𝒑𝒅𝒐𝒑𝒑=𝒎𝒈𝒉

𝒅𝒐𝒑𝒑= .𝟎𝟒𝟖𝒎

𝒅𝒐𝒑𝒑=𝟑𝟎𝟎𝟎𝒌𝒈(𝟏𝟎𝒎

𝒔𝟐 )𝟖𝒎

𝟓𝒙𝟏𝟎𝟔𝑵

A-3: At an amusement park, a ride called the “Cyclone” is a giant roller coaster that ascends a 34.1-m hill and then drops 21.9 m before ascending the next hill. The train of cars has a mass of 4727 kg.

a) How much work is required to get an empty train of cars from the ground to the top of the first hill?

𝒘=𝒎𝒈𝒉

J

(34.1 m)

b) What power must be generated to bring the train to the top of the first hill in 30.0 s?

𝑷 ,𝑾𝒂𝒕𝒕 𝐬=𝒘𝒐𝒓𝒌 , 𝑱𝒕𝒊𝒎𝒆 ,𝒔

Watts

𝑷=𝟏 .𝟔 𝒙𝟏𝟎𝟔 J 𝟑𝟎 𝒔

c) How much PE is converted into KE from the top of the first hill to the bottom of the 21.9 m drop?

)

(21.9 m)

J

A-4 A flea gains 1.0 x 10-7 J of PE jumping up to a height of 0.030 m from a dog’s back. What is the mass of the flea?

∆ 𝑷𝑬=𝐦𝐠𝒉

kg

𝒎=∆𝑷𝑬𝒈𝒉

𝒎= 𝟏 𝒙𝟏𝟎−𝟕 𝑱

(𝟏𝟎𝒎𝒔𝟐 )( .𝟎𝟑𝟎𝒎¿

¿

A-5: At target practice, Diana holds her bow and pulls the arrow back a distance of 0.30 m by exerting an average force of 40.0 N. What is the potential energy stored in the bow the moment before the arrow is released?

𝑷𝑬=𝐅𝐝𝑷𝑬=(𝟒𝟎𝐍 )(𝟎 .𝟑𝐦 )

𝑷𝑬=𝟏𝟐 𝐉

A-6: The coyote, whose mass is 20.0 kg, is chasing the roadrunner when the coyote accidentally runs off the edge of a cliff and plummets to the ground 30.0 m below. What force does the ground exert on the coyote as he makes a coyote-shaped dent 0.420 m deep in the ground?

𝒎𝒈𝒉=𝐅𝐝

𝑭=𝟏𝟒 ,𝟑𝟎𝟎 𝐉

(𝟐𝟎𝒌𝒈 )(𝟏𝟎𝒎𝒔𝟐 )(30 m )=𝐅 (𝟎 .𝟒𝟐𝟎𝐦)

A-7: A 0.080-kg robin, perched on a power line 6.0 m above the ground, swoops down to snatch a worm from the ground and then returns to an 8.0-m-high tree branch with his catch. a) By how much did the bird’s PE increase in its trip from the power line to the tree branch?

∆ 𝑷𝑬=𝒎𝒈∆𝒉

∆ 𝑷𝑬=(𝟎 .𝟎𝟖𝒌𝒈 )(𝟏𝟎 𝒎𝒔𝟐 )( 2 m )

∆ 𝑷𝑬=𝟏.𝟔 𝑱

b) How would your answer have changed if the bird had flown around a bit before landing on the tree branch?

A-8: Cinnamon, whose mass is 5.45 kg, was adopted from a cat shelter and now enjoys napping on top of the refrigerator. Cinnamon rolls over and falls off the refrigerator landing feet first with a KE of 85.5 J as she hits. How tall is the refrigerator?

∆ 𝑷𝑬=∆𝑲 𝑬

∆𝒉=𝟏 .𝟓𝟕𝒎

𝒎𝒈∆𝒉=∆𝑲 𝑬

(𝟓 .𝟒𝟓𝒌𝒈 )(𝟏𝟎𝒎𝒔𝟐 )∆𝒉=𝟖𝟓 .𝟓 𝑱

A-9: Calories measure energy we get from food, and one dietary Calorie is equal to 4187 J. The average food energy intake for human beings is 2000. Calories/day. Assume you have a mass of 55.0 kg and you want to burn off all the Calories you consume in one day. How high a mountain would you have to climb to do so?

∆ 𝑷𝑬=∆𝑲 𝑬4187 J)(One Day)

𝑺𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 ∆𝒉∆𝒉=

(𝟐𝟎𝟎𝟎𝑪𝒂𝒍 )(4187 J)

(𝟓 .𝟒𝟓𝒌𝒈 )(𝟏𝟎𝒎𝒔𝟐 ) 15,200 m


A-10: From a height of 2.15 m above the floor of the basketball court, forward Paul Pierce tosses a shot straight up next to the basketball hoop with a KE of 5.40 J. If his regulation-size basketball has a mass of 0.600 kg, will his shot go as high as the 3.04-m hoop? Use the law of conservation of energy.

∆ 𝑷𝑬=∆𝑲 𝑬 J)𝑺𝒐𝒍𝒗𝒆 𝒇𝒐𝒓 𝒉 𝒇

+ 2.15 m3.07 m


𝐈 𝐭′ 𝐬𝐠𝐨𝐭 𝐭𝐡𝐞 𝐣𝐮𝐢𝐜𝐞 !

A-11: Mr. Macintosh, a computer technician, uses a screwdriver with a handle of radius 1.2 cm to remove a screw in the back of a computer. The screw moves out 0.20 cm on each complete turn. What is the ideal mechanical advantage of the screwdriver?


𝒅𝒐

𝑰𝑴𝑨=𝟏 .𝟐𝒄𝒎𝟎 .𝟐𝟎𝒄𝒎

𝑰𝑴𝑨=𝟔

A-12: Tom’s favorite pastime is fishing. a) How much work is required for Tom to reel in a 10.0-kg bluefish from the water’s surface to the deck of a fishing boat, 5.20 m above the water, if the reel of his fishing pole is 85.0% efficient?

Eff .

𝒘 𝒊=𝟏𝟎𝒌𝒈 (𝟏𝟎𝒎

𝒔𝟐 )𝟓 .𝟐𝒎

𝟎 .𝟖𝟓𝒘 𝒊=𝟔𝟏𝟐 𝑱

b) If Tom applies a force of 15 N to the reel’s crank handle, what is the actual mechanical advantage of the fishing pole?


𝑭 𝒊𝑨𝑴𝑨=

𝟏𝟎𝟎𝑵𝟏𝟓𝑵

𝑨𝑴𝑨=𝟔 .𝟕

c) What is the ideal mechanical advantage of the fishing pole?

𝑬𝒇𝒇 =𝑨𝑴𝑨𝑰𝑴𝑨

𝑰𝑴𝑨=𝑨𝑴𝑨𝑬𝑭𝑭

𝑰𝑴𝑨=𝟔 .𝟕.𝟖𝟓

𝑰𝑴𝑨=𝟕 .𝟗

A -13 A nutcracker 16 cm long is used to crack open a Brazil nut that is placed 12 cm from where your hand is squeezing the nutcracker. What is the ideal mechanical advantage of the nutcracker?


𝒅𝒐

𝑰𝑴𝑨=𝟑

𝑰𝑴𝑨=𝟏𝟐𝒄𝒎𝟒𝒄𝒎

B-1: A 5.00-N salmon swims 20.0 m upstream against a current that provides a resistance of 1.50 N. This portion of the stream rises at an angle of 10.0° with respect to the horizontal. a) How much work is done by the salmon against the current?

𝒘 𝒇 =𝑭 ∆𝒅

B-1: A 5.00-N salmon swims 20.0 m upstream against a current that provides a resistance of 1.50 N. This portion of the stream rises at an angle of 10.0° with respect to the horizontal. b) What is the gain in PE by the salmon?

sinθ)

do

di = 20 m

θ = 100

)

∆ 𝑷𝑬=𝟏𝟕𝟒 𝑱

B-1: A 5.00-N salmon swims 20.0 m upstream against a current that provides a resistance of 1.50 N. This portion of the stream rises at an angle of 10.0° with respect to the horizontal. c) What is the total work that must be done by the salmon?

𝒘 𝑻𝒐𝒕𝒂𝒍=𝟏𝟕 .𝟒 𝑱+𝟑𝟎 𝑱

𝒘 𝑻𝒐𝒕𝒂𝒍=𝟒𝟕 .𝟒 𝑱

𝒘 𝑻𝒐𝒕𝒂𝒍=∆𝑷𝑬+𝒘 𝒇

B-1: A 5.00-N salmon swims 20.0 m upstream against a current that provides a resistance of 1.50 N. This portion of the stream rises at an angle of 10.0° with respect to the horizontal. d) If the salmon takes 40.0 s to swim the distance, what power does it exert in doing so?

𝑷𝒐𝒘𝒆𝒓 ,𝑾=𝑾𝒐𝒓𝒌 , 𝑱𝑻𝒊𝒎𝒆 ,𝒔

𝑷𝒐𝒘𝒆𝒓 ,𝑾=𝟒𝟕 .𝟒 𝑱𝟒𝟎 𝒔 P

B-2: A 30-kg shopping cart full of groceries sitting at the top of a 2.0-m hill begins to roll until it hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart and hits a parked car with an average force of 490 N. How deep a dent is made in the car?

𝑭𝒅=𝒎𝒈𝒉 𝒅=𝒎𝒈𝒉

𝑭

𝒅=.𝟐𝟓𝒌𝒈(𝟏𝟎𝒎

𝒔𝟐 )𝟐𝒎

𝟒𝟗𝟎𝑵𝒅=𝟎 .𝟎𝟏𝟎𝒎

B-3: Using her snowmobile, Midge pulls a 60.0 kg skier up a ski slope inclined at an angle of 12.0° to the horizontal. The snowmobile exerts a force of 200. N parallel to the hill. If the coefficient of friction between the skis and the snow is 0.120, how fast is the skier moving after he has been pulled for 100.0 m starting from rest?

θ = 120

60 kg

Fi = 200 N

μ = 0.12d

i = 200 N do

θ = 120

60 kg

Fi = 200 N

μ = 0.12d

i = 200 N do

V V

V

B-4: Jose, whose mass is 45.0 kg, is riding his 5.0-kg skateboard down the sidewalk with a constant speed of 6.0 m/s when he rolls across a 10.0-m-long patch of sand on the pavement. The sand provides a force of friction of 6.0 N. What is Jose’s speed as he emerges from the sandy section?

𝐯=√(𝒗𝒐 )𝟐−(𝟐𝑭𝒅𝒎 )

v v

B-5: Eben lifts an engine out of his car with the help of a winch that allows him to raise the engine 0.020 m for every 0.90 m he pulls on the cable. Eben expends 1000. J of energy to lift the 800.-N engine 0.50 m. a) What is the efficiency of the winch?

𝑬𝒇𝒇 =𝑭𝒐 𝒅𝒐

𝒘 𝒊

𝑬𝒇𝒇 =(𝟖𝟎𝟎𝑵 )𝟎 .𝟓𝟎𝒎¿ ¿𝟏𝟎𝟎𝟎 𝑱

𝑬𝒇𝒇 =𝟎 .𝟒

B-5: Eben lifts an engine out of his car with the help of a winch that allows him to raise the engine 0.020 m for every 0.90 m he pulls on the cable. Eben expends 1000. J of energy to lift the 800.-N engine 0.50 m. b) What is the ideal mechanical advantage of the winch?

𝐈𝐌𝐀=𝒅𝒊

𝒅𝒐𝐈𝐌𝐀=

𝟎 .𝟗𝒎𝟎 .𝟎𝟐𝒎

𝐈𝐌𝐀=𝟒𝟓

B-5: Eben lifts an engine out of his car with the help of a winch that allows him to raise the engine 0.020 m for every 0.90 m he pulls on the cable. Eben expends 1000. J of energy to lift the 800.-N engine 0.50 m. c) What is the actual mechanical advantage of the winch?

𝐀𝐌𝐀=(𝑰𝑴𝑨)(𝑬𝒇𝒇 )𝐀𝐌𝐀=(𝟒𝟓)(𝟎 .𝟒)

𝐀𝐌𝐀=𝟏𝟖

B-5: Eben lifts an engine out of his car with the help of a winch that allows him to raise the engine 0.020 m for every 0.90 m he pulls on the cable. Eben expends 1000. J of energy to lift the 800.-N engine 0.50 m. d) What force does Eben exert to lift the engine?

𝑭 𝒊𝒏=𝑭𝒐𝒖𝒕

𝑨𝑴𝑨𝑭 𝒊𝒏=

𝟖𝟎𝟎𝑵𝟏𝟖𝑭 𝒊𝒏=𝟒𝟒𝑵

Book

Problems

1. A force sets an object in motion. When the force is multiplied by the time of its application, we call the quantity impulse, which changes the momentum of that object. What do we call the quantity force distance, and what quantity can this change?2. Work is required to lift a barbell. How many times more work is required to lift the barbell three times as high?3. Which requires more work, lifting a 10-kg load a vertical distance of 2 m or lifting a 5-kg load a vertical distance of 4 m?1. Work; Object’s Energy2. Three3. Both the same, 200 J

4. How many joules of work are done on an object when a force of 10 N pushes it a distance of 10 m?5. How much power is required to do 100 J of work on an object in a time of 0.5 s? How much power is required if the same work is done in 1 s?6. What are the two main forms of mechanical energy?4. 5.

6. PE and KE

7. a. If you do 100 J of work to elevate a bucket of water, what is its gravitational potential energy relative to its starting position?b. What would the gravitational potential energy be if the bucket were raised twice as high?8. A boulder is raised above the ground so that its potential energy relative to the ground is 200 J. Then it is dropped. What is its kinetic energy just before it hits the ground?7. a. 100 J b. 200 J8. 200 J

9. Suppose you know the amount of work the brakes of a car must do to stop a car at a given speed. How much work must they do to stop a car that is moving four times as fast? How will the stopping distances compare?10. How does speed affect the friction between a road and a skidding tire?11. What will be the kinetic energy of an arrow having a potential energy of 50 J after it is shot from a bow? 9. 16 times as much work 16 times the distance10. Speed does not affect friction.11. 50 J

12. What does it mean to say that in any system the total energy score stays the same?13. In what sense is energy from coal actually solar energy?14. How does the amount of work done on an automobile by its engine relate to the energy content of the gasoline?15. In what two ways can a machine alter an input force?12. Energy is conserved.13. Material that forms coal was produced by sun’s energy.14. It is less than the energy in the gasoline.15. It can change its magnitude and/or direction

16. In what way is a machine subject to the law of energy conservation? Is it possible for a machine to multiply energy or work input?17. What does it mean to say that a machine has a certain mechanical advantage? 18. In which type of lever is the output force smaller than the input force?19. What is the efficiency of a machine that requires 100 J of energy to do 35 J of work?16. Work out cannot exceed work in; no.17. It can multiply force by a certain amount.18. Type 3—always, Type 1— maybe19. 35%

20. Distinguish between theoretical mechanical advantage and actual mechanical advantage. How would these compare if a machine were 100% efficient?21. What is the efficiency of her body when a cyclist expends 1000 W of power to deliver mechanical energy to the bicycle at the rate of 100 W?22. In what sense are our bodies machines?20. TMA—no friction AMA—with friction; same21. 10%22. Like machines, our bodies need an energy supply. Also, the same principle of combustion occurs in the metabolism of food in the body and the burning of fossil fuels.

23. What is the ultimate source of the energy derived from the burning of fossil fuels, from dams, and from windmills?24. What is the ultimate source of geothermal energy?25. Can we correctly say that a new source of energy is hydrogen? Why or why not?23. The sun24. Radioactivity in Earth’s interior25. No. Hydrogen is not a new source of energy because it takes energy to extract hydrogen from water and carbon compounds.

26. The mass and speed of three vehicles are shown below.a. Rank the vehicles by momentum from greatest to least.b. Rank the vehicles by kinetic energy from greatest to least.26. a. B, A, C b. C, B, A

27. Consider these four situations.(A) a 3-kg ball at rest atop a 5-m-tall hill(B) a 4-kg ball at rest atop a 5-m-tall hill(C) a 3-kg ball moving at 2 m/s atop a 5-m-tall hill(D) a 4-kg ball moving at 2 m/s at ground levela. Rank from greatest to least the potential energy of each ball.b. Rank from greatest to least the kinetic energy of each ball.c. Rank from greatest to least the total energy of each ball.27. a. B, A = C, D b. D, C, A = B c. B, C, A, D

28. A ball is released at the left end of the metal track shown below. Assume it has only enough friction to roll, but not to lessen its speed.a. Rank from greatest to least the ball’s momentum at each point.b. Rank from greatest to least the ball’s kinetic energy at each point.c. Rank from greatest to least the ball’s potential energy at each point.

28. a. C, B = D, A b. C, B = D, A c. A, B = D, C

29. The roller coaster ride starts with the car at rest at point A.a. Rank from greatest to least the car’s speed at each point.b. Rank from greatest to least the car’s kinetic energy at each point.c. Rank from greatest to least the car’s potential energy at each point.29. a. D, B, C, E, A b. D, B, C, E, A c. A, E, C, B, D

30. Rank the efficiency of these machines from highest to lowest.(A) energy in 100 J; energy out 60 J(B) energy in 100 J; energy out 50 J(C) energy in 200 J; energy out 80 J(D) energy in 200 J; energy out 120 J30. A = D, B, C

31. Carts moving along the lab floor run up short inclines. Friction effects are negligible.a. Rank the carts by kinetic energy before they meet the incline.b. Rank the carts by how high they go up the incline.c. Rank the carts by potential energy when they reach the highest point on the incline.d. Why are your answers different for b and c?31. a. B, A, C, D b. A, B, C, D c. B, A, C, D d. PETop = KEBottom

32. Rank the scale readings from greatest to least. (Ignore friction.) 33. Calculate the work done when a force of 1 N moves a book 2 m.34. Calculate the work done when a 20-N force pushes a cart 3.5 m.35. Calculate the work done in lifting a 500-N barbell 2.2 m above the floor. (What is the potential energy of the barbell when it is lifted to this height32. 33. 34. 35.

36. Calculate the watts of power expended when a force of 1 N moves a book 2 m in a time interval of 1 s.37. Calculate the power expended when a 20-N force pushes a cart 3.5 m in a time of 0.5 s.

38. How many joules of potential energy does a 1-kg book gain when it is elevated 4 m? When it is elevated 8 m? 36. 37. 38. = (1 kg)(10 N/kg)(4 m) = 40 J =80 J

39. Calculate the increase in potential energy when a 20-kg block of ice is lifted a vertical distance of 2 m.40. Calculate the number of joules of kinetic energy a 1-kg book has when tossed across the room at a speed of 2 m/s.41. How much work is required to increase the kinetic energy of a car by 5000 J?42. What change in kinetic energy does an airplane experience on takeoff if it is moved a distance of 500 m by a sustained net force of 5000 N?39. 40. 41. 42.

43. Which requires more work: stretching a strong spring a certain distance or stretching a weak spring the same distance? 44. Two people who weigh the same amount climb a flight of stairs. The first person climbs the stairs in 30 s, while the second person climbs them in 40 s. Which person does more work? Which uses more power?43. More force to stretch strong spring, so more work in stretching the same distance.44. Same work done by each, for same hour; climber in 30 s uses more power due to shorter time.

45. A physics teacher demonstrates energy conservation by releasing a heavy pendulum bob, as shown in the sketch, allowing it to swing to and fro. What would happen if, in his exuberance, he gave the bob a slight shove as it left his nose? Explain.46. Consider the kinetic energy of a fly in the cabin of a fast-moving train. Does it have the same or different kinetic energies relative to the train? Relative to the ground outside?45. If ball is given an initial KE, it returns to its starting position with that KE (moving in the other direction!) and hits the instructor.46. Just as motion is relative, KE is also. The speed and KE of the fly are different relative to the train and the ground.

47. When a driver applies brakes to keep a car going downhill at constant speed and constant kinetic energy, the potential energy of the car decreases. Where does this energy go? Where does most of it go in a hybrid vehicle?48. What is the theoretical mechanical advantage for each of the three lever systems shown?47. Energy is wasted as heat in a non-hybrid car. In a hybrid car, energy charges batteries and is converted to electricity.48. 1 2 0.5

49. Dry-rock geothermal power can be a major contributor to power with no pollution. The bottom of a hole drilled down into Earth’s interior is fractured, making a large surfaced hot cavity. Water is introduced from the top by a second hole. Superheated water rising to the surface then drives a conventional turbine to produce electricity. What is the source of this energy?50. A stuntman on a cliff has a PE of 10,000 J. Show that when his potential energy is 2000 J, his kinetic energy is 8000 J.49. Energy from radioactive decay in Earth’s interior50. KE =

51. Relative to the ground below, how many joules of PE does a 1000-N boulder have at the top of a 5-m ledge? If it falls, with how much KE will it strike the ground? What will be its speed on impact?

52. A hammer falls off a rooftop and strikes the ground with a certain KE. If it fell from a roof that was four times higher, how would its KE of impact compare? Its speed of impact? (Neglect air resistance.)Four times higher means four times the PE, thus four times the KE of impact. From

This means twice the impact speed (because ). This result can be obtained from

…where falling from 4d takes twice the time. Twice the time at the same acceleration g means twice the speed; .

53. A car can go from 0 to 100 km/h in 10 s. If the engine delivered twice the power, how many seconds would it take?54. If a car traveling at 60 km/h will skid 20 m when its brakes lock, how far will it skid if it is traveling at 120 km/h when its brakes lock? (This question is typical on some driver’s license exams.)53. Twice the power means doing twice the work in the same time or the same work in half the time. To achieve the same change in speed (and same change in KE) with twice the power means the work can be done in half the time, or 5 s.54. Twice the speed means four times the KE, and four times the work to reduce the KE to zero. F is constant so d = 80 m.

Additional

Practice Problem

s

1. A skater with a mass of 52.0 kg moving at 2.5 m/s glides to a stop over a distance of 24.0 m. How much work did the friction of the ice do to bring the skater to a stop? How much work would the skater have to do to speed up to 2.5 m/s again? 𝐖=𝑲𝑬 𝒇 −𝑲𝑬𝒐

𝐖=+𝟏𝟔𝟎 𝑱

𝐖=𝟏𝟐𝐦𝒗 𝒇

𝟐−𝟏𝟐𝐦𝒗𝒐

𝟐

𝐖=𝟏𝟐

(𝟓𝟐𝐤𝐠 )(𝟎 𝒎𝒔 )

𝟐

−𝟏𝟐

(𝟓𝟐𝐤𝐠)(𝟐 .𝟓𝒎𝒔 )

𝟐

2. An 875.0-kg compact car speeds up from 22.0 m/s to 44.0 m/s while passing another car. What are its initial and final energies, and how much work is done on the car to increase its speed? 𝐖=𝑲𝑬 𝒇 −𝑲𝑬𝒐

𝐖=+𝟔𝟑𝟓 ,𝟎𝟎𝟎 𝑱

𝐖=𝟏𝟐𝐦𝒗 𝒇


𝟐

𝐖=𝟏𝟐

(𝟖𝟕𝟓𝐤𝐠 )(𝟒𝟒 𝒎𝒔 )

𝟐

−𝟏𝟐

(𝟖𝟕𝟓𝐤𝐠 )(𝟐𝟐𝒎𝒔 )

𝟐

3. A comet with a mass of 7.85 x1011 kg strikes Earth at a speed of 25.0 km/s. Find kinetic energy of the comet in joules, and compare the work that is done by Earth in stopping the comet to the 4.2x 1015 J of energy that was released by the largest nuclear weapon ever built.

𝑲𝑬 𝒄𝒐𝒎𝒆𝒕=𝟐 .𝟒𝟓𝐱𝟏𝟎𝟐𝟎 𝑱

𝐖=𝟏𝟐𝐦𝒗𝟐

𝑲𝑬 𝒄𝒐𝒎𝒆𝒕=𝟏𝟐

(𝟕 .𝟖𝟓𝐱𝟏𝟎𝟏𝟏𝐤𝐠 )(𝟐 .𝟓𝐱𝟏𝟎𝟒 𝒎𝒔 )

𝟐

5.8 x 104 bombs would be required to produce the same amount of energy used by Earth in stopping the comet.

𝑲𝑬𝒄𝒐𝒎𝒆𝒕

𝑲𝑬𝒃𝒐𝒎𝒃

=𝟐 .𝟒𝟓𝐱𝟏𝟎𝟐𝟎 𝑱 𝟒 .𝟐𝐱 𝟏𝟎𝟏𝟓 𝑱

=𝟓 .𝟖𝐱𝟏𝟎𝟒 𝑱

4. In Example Problem 1, what is the potential energy of the bowling ball relative to the rack when it is on the floor?

PE =mgh PE = (7.30 kg)(9.80 m/s2)(-0.610 m)

PE =-43.6 J5. If you slowly lower a 20.0 kg bag of sand 1.20 m from the trunk of a car to the driveway, how much work do you do?

W = Fd W = mg(hf hi)

W = (20.0 kg)(9.80 m/s2)(0.00 m - 1.20 m)

W = 2.35x102 J

6. A boy lifts a 2.2-kg book from his desk, which is 0.80 m high, to a bookshelf that is 2.10 m high. What is the potential energy of the book relative to the desk?

7. If a 1.8-kg brick falls to the ground from a chimney that is 6.7 m high, what is the change in its potential energy?

𝑷𝑬=𝒎𝒈 (𝒉 𝒇−𝒉 𝒇 )𝑷𝑬=(𝟐 .𝟐𝒌𝒈 )(𝟗 .𝟖𝒎

𝒔𝟐) (𝟐 .𝟏𝒎−𝟎.𝟖𝐦 )𝑷𝑬=𝟐𝟖 𝑱

𝑷𝑬=𝒎𝒈 (𝒉 𝒇−𝒉 𝒇 )𝑷𝑬=𝒎𝐠 (𝒉 𝒇−𝒉 𝒇 )𝑷𝑬=(𝟏 .𝟖𝒌𝒈 )(𝟗 .𝟖𝒎

𝒔𝟐 )(𝟎𝒎−𝟔 .𝟕𝐦 )

𝑷𝑬=−𝟏𝟐𝟎 𝑱

8. A warehouse worker picks up a 10.1-kg box from the floor and sets it on a long, 1.1-m-high table. He slides the box 5.0 m along the table and then lowers it back to the floor. What were the changes in the energy of the box, and how did the total energy of the box change? (Ignore friction.)

To slide the box across the table, W 0.0 because the height did not change and we ignored friction. To lower the box to the floor:

∆ 𝑷𝑬= (𝟏𝟎 .𝟏𝒌𝒈 )(𝟗 .𝟖𝒎𝒔𝟐 ) (𝟏.𝟏𝒎−𝟎𝐦 )=𝟏𝟏𝟎 𝐉

𝑷𝑬=(𝟏𝟎 .𝟏𝒌𝒈 )(𝟗 .𝟖𝒎𝒔𝟐) (𝟎𝒎−𝟏 .𝟏𝐦 )=−𝟏𝟏𝟎 𝐉

𝑾 =𝑭𝒅=𝑷𝑬=𝒎𝒈 (𝒉𝒇 −𝒉𝒇 )=∆𝑷𝑬


8. A warehouse worker picks up a 10.1-kg box from the floor and sets it on a long, 1.1-m-high table. He slides the box 5.0 m along the table and then lowers it back to the floor. What were the changes in the energy of the box, and how did the total energy of the box change? (Ignore friction.)

𝑷𝑬=(𝟏𝟎 .𝟏𝒌𝒈 )(𝟗 .𝟖𝒎𝒔𝟐) (𝟎𝒎−𝟏 .𝟏𝐦 )=−𝟏𝟏𝟎 𝐉


The sum of the three energy changes is…𝟏𝟏𝟎 𝑱+𝟎 𝑱 +(−𝟏𝟏𝟎 𝑱 )=𝟎 𝑱

9. Elastic Potential Energy You get a spring loaded toy pistol ready to fire by compressing the spring. The elastic potential energy of the spring pushes the rubber dart out of the pistol. You use the toy pistol to shoot the dart straight up. Draw bar graphs that describe the forms of energy present in the following instances. a. The dart is pushed into the gun barrel, thereby compressing the spring.There should be three bars:

one for the spring’s potential energy, one for gravitational potential energy, and one for kinetic energy. The spring’s potential energy is at the maximum level, and the other two are zero.

b. The spring expands and the dart leaves the gun barrel after the trigger is pulled.

The kinetic energy is at the maximum level, and the other two are zero.

c. The dart reaches the top of its flight.

The gravitational potential energy is at the maximum level, and the other two are zero.

10. Potential Energy A 25.0-kg shell is shot from a cannon at Earth’s surface. The reference level is Earth’s surface. What is the gravitational potential energy of the system when the shell is at 425 m? What is the change in potential energy when the shell falls to a height of 225 m?

𝑷𝑬=(𝟐𝟓𝒌𝒈 )(𝟗 .𝟖 𝒎𝒔𝟐 ) (𝟒𝟐𝟓𝒎 )=𝟏𝟎𝟒 ,𝟎𝟎𝟎 𝐉

𝟏𝟎𝟒 ,𝟎𝟎𝟎 𝑱 −𝟓𝟓 ,𝟏𝟎𝟎 𝑱=𝟒𝟖 ,𝟗𝟎𝟎 𝑱

𝒂¿𝑷𝑬=𝒎𝒈𝒉𝒃¿ 𝑷𝑬=𝒎𝒈𝒉

𝑷𝑬=(𝟐𝟓𝒌𝒈 )(𝟗 .𝟖 𝒎𝒔𝟐 ) (𝟐𝟐𝟓𝒎 )=𝟓𝟓 ,𝟏𝟎𝟎 𝐉

The change in energy is…

11. Rotational Kinetic Energy Suppose some children push a merry-go-round so that it turns twice as fast as it did before they pushed it. What are the relative changes in angular momentum and rotational kinetic energy?The angular momentum is doubled because it is proportional to the angular velocity. The rotational kinetic energy is quadrupled because it is proportional to the square of the angular velocity. The children did work in rotating the merry-go-round.

12. Work-Energy Theorem How can you apply the work-energy theorem to lifting a bowling ball from a storage rack to your shoulder?The bowling ball has zero kinetic energy when it is resting on the rack or when it is held near your shoulder. Therefore, the total work done on the ball by you and by gravity must equal zero.

13. Potential Energy A 90.0-kg rock climber first climbs 45.0 m up to the top of a quarry, then descends 85.0 m from the top to the bottom of the quarry. If the initial height is the reference level, find the potential energy of the system (the climber and Earth) at the top and at the bottom. Draw bar graphs for both situations.

𝑨𝒕𝑻𝒐𝒑 𝑷𝑬=(𝟗𝟎𝒌𝒈 )(𝟗 .𝟖𝒎𝒔𝟐) (+𝟒𝟓𝒎 )=𝟑𝟗 ,𝟕𝟎𝟎 𝐉

𝑷𝑬=𝒎𝒈𝒉𝑨𝒕𝑻𝒐𝒑 𝑷𝑬=𝟑𝟗 ,𝟕𝟎𝟎 𝐉

𝑨𝒕 𝑩𝒐𝒕𝒕𝒐𝒎𝑷𝑬=(𝟗𝟎𝒌𝒈)(𝟗.𝟖 𝒎𝒔𝟐 ) (−𝟒𝟎𝒎 )=𝟑𝟓 ,𝟑𝟎𝟎 𝐉

14. Karl uses an air hose to exert a constant horizontal force on a puck, which is on a frictionless air table. He keeps the hose aimed at the puck, thereby creating a constant force as the puck moves a fixed distance.a. Explain what happens in terms of work and energy. Draw bar graphs. Karl exerted a constant force F over a distance d and did an amount of work W =Fd on the puck. This work changed the kinetic energy of the puck.

b. Suppose Karl uses a different puck with half the mass of the first one. All other conditions remain the same. How will the kinetic energy and work differ from those in the first situation?If the puck has half the mass, it still receives the same amount of work and has the same change in kinetic energy. However, the smaller mass will move faster by a factor of 1.414.

𝐖=𝑲𝑬 𝒇 −𝑲𝑬𝒐𝐖=𝟏𝟐𝐦𝒗 𝒇


𝟐𝐖=𝟏𝟐𝐦𝒗 𝒇

𝟐

c. Explain what happened in parts a and b in terms of impulse and momentum. The two pucks do not have the same final momentum. Momentum of the first puck:

P1 = m1 + v1

Momentum of the second puck:P2 = m2v2

Thus, the second puck has less momentum than the first puck does. Because the change in momentum is equal to the impulse provided by the air hose, the second puck receives a smaller impulse.

𝐖=(𝟏𝟐𝒎𝟏)(𝟏 .𝟒𝟏𝟒𝒗¿¿𝟏)= .𝟕𝟎𝟕𝒑𝟏¿

15. A bike rider approaches a hill at a speed of 8.5 m/s. The combined mass of the bike and the rider is 85.0 kg. Choose a suitable system. Find the initial kinetic energy of the system. The rider coasts up the hill. Assuming there is no friction, at what height will the bike come to rest?The system is the (bike + rider + Earth). There are no external forces, so total energy is conserved.

𝑲𝑬 𝒐+𝑷𝑬𝒐=𝑲𝑬 𝒇 +𝑷𝑬 𝒇

𝐊𝐄=𝟏𝟐𝐦 𝒗𝟐𝐊𝐄=𝟏

𝟐(𝟖𝟓𝐤𝐠)(𝟖 .𝟓 𝒎

𝒔 )𝟐𝐊𝐄=𝟑 ,𝟏𝟎𝟎 𝑱

𝟏𝟐𝐦 𝒗𝟐+𝟎=𝟎+𝐦𝐠𝐡

𝒉= 𝒗𝟐

𝟐𝒈𝒉=

(𝟖 .𝟓𝒎𝒔 )

𝟐

𝟐(𝟗 .𝟖 𝒎𝒔𝟐 )

𝒉=𝟑.𝟕𝒎

16. Suppose that the bike rider in problem 15 pedaled up the hill and never came to a stop. In what system is energy conserved? From what form of energy did the bike gain mechanical energy?The system of Earth, bike, and rider remains the same, but now the energy involved is not mechanical energy alone. The rider must be considered as having stored energy, some of which is converted to mechanical energy. Energy came from the chemical potential energy stored in the rider’s body.

17. A skier starts from rest at the top of a 45.0-m-high hill, skis down a 30° incline into a valley, and continues up a 40.0-mhigh hill. The heights of both hills are measured from the valley floor. Assume that you can neglect friction and the effect of the ski poles. How fast is the skier moving at the bottom of the valley? What is the skier’s speed at the top of the next hill? Do the angles of the hills affect your answers?

𝑲𝑬 𝒐+𝑷𝑬𝒐=𝑲𝑬 𝒇 +𝑷𝑬 𝒇𝑩𝒐𝒕𝒕𝒐𝒎

𝟎+𝐦𝐠𝐡=𝟏𝟐𝐦 𝒗𝟐+𝟎

𝒗𝟐=𝟐𝒈𝒉

𝒗=𝟐𝟗 .𝟕𝒎𝒔

𝒗=√𝟐𝒈𝒉𝒗=√𝟐 (𝟗 .𝟖 𝒎

𝒔𝟐 )(𝟒𝟓𝒎)

𝑲𝑬 𝒐+𝑷𝑬𝒐=𝑲𝑬 𝒇 +𝑷𝑬 𝒇

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𝟎+𝐦𝐠𝒉𝒐=𝟏𝟐𝐦𝒗𝟐+𝐦𝐠𝒉 𝒇𝒗𝟐=𝟐𝒈 (𝒉¿¿𝟎−𝒉 𝒇 )¿

𝒗=𝟗 .𝟗𝒎𝒔 𝒗=√𝟐𝒈(𝒉¿¿𝟎−𝒉 𝒇 ) ¿

𝒗=√𝟐 (𝟗 .𝟖 𝒎𝒔𝟐 )(𝟒𝟓𝒎−𝟒𝟎𝒎)

No, the angles do not have any impact.

18. In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. The size of the splash depends not only on the diver’s style, but also on the amount of kinetic energy that the diver has. Consider a contest in which each diver jumps from a 3.00-m platform. One diver has a mass of 136 kg and simply steps off the platform. Another diver has a mass of 102 kg and leaps upward from the platform. How high would the second diver have to leap to make a competitive splash?Using the water as a reference level, the kinetic energy on entry is equal to the potential energy of the diver at the top of his flight.

Using the water as a reference level, the kinetic energy on entry is equal to the potential energy of the diver at the top of his flight. The large diver has… 𝐏𝐄=𝐦𝐠𝐡

𝑷𝑬=(𝟏𝟑𝟔𝒌𝒈)(𝟗 .𝟖𝒎𝒔𝟐 ) (𝟑𝒎 )=𝟒𝟎𝟎𝟎 𝑱

𝒉=𝟒𝟎𝟎𝟎 𝑱

(𝟏𝟎𝟐𝒌𝒈)(𝟗 .𝟖𝒎𝒔𝟐 )

=𝟒𝐦

19. An 8.00-g bullet is fired horizontally into a 9.00-kg block of wood on an air table and is embedded in it. After the collision, the block and bullet slide along the frictionless surface together with a speed of 10.0 cm/s. What was the initial speed of the bullet?

𝒎𝒗=(𝐦+𝐌 )𝐕𝒐𝒓

𝒗=(𝐦+𝐌 )𝐕

𝒎

𝒗=(𝟎 .𝟎𝟎𝟖𝟎𝐤𝐠+𝟗𝐤𝐠 )(𝟎 .𝟏 𝒎

𝒔 )

𝟎 .𝟎𝟎𝟖𝟎𝒌𝒈

𝒗=𝟏𝟏𝟑𝒎𝒔

20. A 0.73-kg magnetic target is suspended on a string. A 0.025-kg magnetic dart, shot horizontally, strikes the target head-on. The dart and the target together, acting like a pendulum, swing 12.0 cm above the initial level before instantaneously coming to rest.a. Sketch the situation and choose a system.

The system includes the suspended target and the dart.

b. Decide what is conserved in each part and explain your decision. Only momentum is conserved in the inelastic dart-target collision, so

mvo + MVo = (m + M)Vf

Where, Vi = 0 since the target is initially at rest and Vf is the common velocity just after impact. As the dart-target combination swings upward, energy is conserved, so

ΔPE = ΔKE or, at the top of the swing,

(𝐦+𝐌 )𝐠𝒉 𝒇=𝟏𝟐

(𝐦+𝐌) (𝑽 𝒇 )𝟐

b. What was the initial velocity of the dart?Solve for Vf …

𝑽 𝒊=𝟒𝟔𝒎𝒔

𝑽 𝒇=√𝟐𝒈𝒉𝒇

Substitute vf into the momentum equation and solve for vi.

𝑽 𝒊=(𝟎 .𝟎𝟐𝟓𝒌𝒈+𝟎 .𝟕𝟑𝒌𝒈𝟎 .𝟎𝟐𝟓𝒌𝒈 )√𝟐 (𝟗 .𝟖 𝒎

𝒔𝟐 )(𝟎 .𝟏𝟐𝟎 )

21. A 91.0 kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.1 m/s in the same direction, hits him from behind. They slide off together.

a. What are the total energy and momentum in the system before the collision?

𝑲𝑬 𝒊=𝟏𝟐

𝒎𝟏(𝒗 ¿¿𝟏)𝟐+𝟏𝟐

𝒎𝟐(𝒗¿¿𝟐)𝟐 ¿¿

𝑲𝑬 𝒊=𝟏𝟐𝟗𝟏𝒌𝒈 (𝟓 .𝟓𝒎

𝒔 )𝟐

+𝟏𝟐𝟗𝟏𝒌𝒈(𝟖 .𝟏 𝒎

𝒔 )𝟐

𝑷 𝒊=𝒎𝟏𝒗𝟏+𝒎𝟐𝒗𝟐

𝑷 𝒊=(𝟗𝟏𝒌𝒈 )(𝟓 .𝟓𝒎𝒔 )+(𝟗𝟏𝒌𝒈 )(𝟖 .𝟏𝒎

𝒔 )𝑷 𝒊=(𝟏 ,𝟐𝟎𝟎 𝒌𝒈𝒎

𝒔 )

𝑲𝑬 𝒊=𝟒 ,𝟒𝟎𝟎 𝑱

21. A 91.0 kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.1 m/s in the same direction, hits him from behind. They slide off together.

b. What is the velocity of the two hockey players after the collision?

𝑷 𝒊=𝒎𝟏𝒗𝟏+𝒎𝟐𝒗𝟐 𝒗 𝒇=𝟔.𝟖𝒎𝒔

𝒎𝟏𝒗𝟏+𝒎𝟐𝒗𝟐=¿ ¿

𝑷 𝒊=𝑷 𝒇

𝒗 𝒇=𝒎𝟏𝒗𝟏+𝒎𝟐𝒗𝟐

𝒎𝟏+𝒎𝟐

𝒗 𝒇=(𝟗𝟏𝒌𝒈)(𝟓 .𝟓 𝒎

𝒔 )+(𝟗𝟏𝒌𝒈)(𝟖.𝟏𝒎𝒔 )

𝟏𝟖𝟐𝒌𝒈

21. A 91.0 kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.1 m/s in the same direction, hits him from behind. They slide off together.c. How much energy was lost in the collision?The KEf is:

𝑲𝑬 𝒇=𝟏𝟐

¿ ¿

𝑲𝑬 𝒇=𝟏𝟐

(𝟗𝟏𝒌𝒈+𝟗𝟏𝒌𝒈)(𝟔 .𝟖 𝒎𝒔 )

𝟐

𝑲𝑬 𝒇=𝟒 ,𝟐𝟎𝟎 𝑱∆ 𝑲𝑬=𝑲𝑬 𝒇 −𝑲𝑬 𝒐=𝟒 ,𝟒𝟎𝟎 𝑱 −𝟒 ,𝟐𝟎𝟎 𝑱

∆ 𝑲𝑬=𝟐𝟎𝟎 𝑱

lecture 07 w, e and sm's (1)

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