Physics P Review 4 Energy and Its Conservation

Review 4 Energy and Its Conservation

1. A player pushes a 250–g hockey puck over frictionless ice with a constant force, causing it to accelerate at 24 m/s2 over a distance of 50 cm. a. Find the work done by the hockey

player on the puck. b. What is the change in the kinetic

energy of the puck?

2. You exert a horizontal force of 4.6 N on a textbook as you slide it 0.6 m across a library table to a friend. Calculate the work you do on the book.

3. An electric motor lifts an elevator at a constant speed of 15 m/s. The engine must exert a force of 9000 N in order to balance the weight of the elevator and the friction in the elevator cable. What power does the motor produce in kW?

4. Leah is helping to build a water habitat in a neighborhood park. The habitat includes an upper pond connected to a lower pond, 3.2 m below, by a trickling stream with several small cascades. At a home-building store, she finds a 45–W pump that has a maximum circulation rate of 1900 L of water per hour. Can the pump develop enough power to raise the water from the lower to the upper pond? (The mass density of water, ρ, is 1.00 kg/L.)

5. A gardener lifts a 25–kg bag of sand to a height of 1.1 m, carries it across the yard a distance of 15 m and sets it down against the wall. a. How much work does the gardener

do when he lifts the bag of sand? b. How much total work is done after

the gardener sets down the bag of sand?

6. A 0.149–kg baseball is thrown at a speed of 6.5 m/s. The batter hits the ball and it flies into the outfield at a speed of 19.2 m/s. How much work is done on the baseball?

7. A 6–g block initially at rest is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 1.2 × 10–2 N for a distance of 3 cm. a. What is the work done by the force? b. What is the change in the kinetic

energy of the block? c. What is the speed of the block after

the force is removed?

8. Zeke slides down a snow hill on a rubber mat. Zeke’s mass is 76 kg and the mass of the mat is 2 kg. Zeke starts from rest at the crest of the hill. You may ignore friction. a. What is the change in the

gravitational potential energy of Zeke and the mat when they slide to 1.2 m below the crest?

b. What is the change in the kinetic energy of Zeke and the mat when they slide to 1.2 m below the crest?

c. How fast are Zeke and the mat moving when they are 1.2 m below the crest?

9. Kuan stands on the edge of a building’s roof, 12 m above the ground, and throws a 149–g baseball straight down. The ball hits the ground at a speed of 18 m/s. What was the initial speed of the ball?

10. Meena releases her 10.5–kg toboggan from rest on a hill. The toboggan glides down the frictionless slope of the hill, and at the bottom of the slope it moves along a rough horizontal surface, which exerts a constant frictional force on the toboggan. a. When the toboggan is released from

a height of 15 m, it travels 6 m along the horizontal surface before coming to rest. How much work does the frictional force do on the toboggan?

b. From what height should the toboggan be released so that it stops after traveling 10 m on the horizontal surface?

Physics P Review 4 Energy and Its Conservation

1a. W =Fd W =mad

W = 0.25%kg( ) 24%m/s2( ) 0.5%m( )

W =3"J 1b. W = ΔKE ΔKE=3%J 2. W =Fd W = 4.6$N( ) 0.6$m( )

W =2.76%J

3. P =Wt

P = Fdt

P =Fv P = 9000#N( ) 15#m/s( )

P =135$000$W$=$135$kW

4. P =Wt

P =ΔPEGt

P =mght

P =1900$kg( )g 3.2$m( )

3600$s

P =16.6$W Since the pond only requires 16.6 W the

45 W pump will suffice. 5a. W = ΔPEG W =mgh

W = 25#kg( )g 1.1#m( )

W =270$J 5b. W = ΔPEG +ΔKE W =0

6. W = ΔKE

W =KEf −KEi W = 1

2mvf2 − 12mvi2

W = 12m vf2 −v i2( )

W = 12 0.149'kg( ) 19.2'm/s( )2 − 6.5'm/s( )2⎡

⎣⎢⎤⎦⎥

W =24#J 7a. W =Fd W = 0.012%N( ) 0.03%m( )

W =0.00036%J 7b. W = ΔKE

ΔKE=0.000&36&J

7c. KE = 12mv 2

v = 2KEm

v =2 0.000$36$J( )0.006$kg

v =0.35%m/s 8a. ΔPEG =PEf −PEi

ΔPEG =mghf −mghi

ΔPEG =0− 76$kg$+$2$kg( )g 1.2$m( )

ΔPEG = !920%J 8b. ΔPEG +ΔKE =0

ΔKE = −ΔPEG

ΔKE = 920$J

Physics P Review 4 Energy and Its Conservation

8c. KE = 12mv 2

v = 2KEm

v =2 920$J( )

76$kg$+$2$kg

v = 4.8$m/s 9. ΔPEG +ΔKE =0

PEf −PEi +KEf −KEi =0

0−mghi + 12mvf2 − 1

2mvi2 =0

−ghi + 12v f2 − 1

2v i2 =0

v i = v f2 −2ghi

v i = 18#m/s( )2 −2g 12#m( )

v i = 9.4$m/s 10a. The toboggan starts with PEG, that gets

converted completely into KE, which then becomes heat as the toboggan slides to rest.

W = ΔPEG Fd =mgh

F =mghd

F =10.5%kg( )g 15%m( )

6%m

F =260$N

10b. F =mghd

h = Fdmg

h = 260$N( ) 10$m( )10.5$kg( )g

h =25#m