force on a spring lab… the force applied to a spring and the amount of stretch in the spring are...

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Force on a spring lab… • The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known as Hooke’s Law s Slope = spring constant , k – describes the stiffness of the spring Area under line = stored E el ½ bh = ½ (s)(F) = ½ (s)(ks) = ½ ks 2

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Page 1: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Force on a spring lab…• The force applied to a spring and the amount of stretch

in the spring are directly proportional…

F = ks FThis relationship is known as Hooke’s Law

sSlope = spring constant, k – describes the stiffness of the spring

Area under line = stored Eel

½ bh = ½ (s)(F) = ½ (s)(ks) = ½ ks2

Page 2: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Now, let’s look at Gravitational Potential Energy in a little more detail…

• Construct a free body diagram for a box being lifted at a constant speed and write a force equation.

Page 3: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Now, let’s look at Gravitational Potential Energy in a little more detail…

• Construct a free body diagram for a box being lifted at a constant speed and write a force equation

F

Fg

Since speed is constant, F = 0…F – Fg = 0F – mg = 0F = mgSo, the lifting force is equal to the weight of the box throughout the entire lift.

Page 4: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Gravitational Potential Energy

• In the spring lab we graphed Force vs. Stretch to inform us about stored energy. What would a graph of Force vs. Height look like for the case of lifting the box at constant speed?

Page 5: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Gravitational Potential Energy

• In the spring lab we graphed Force vs. Stretch to inform us about stored energy. What would a graph of Force vs. Height look like for the case of lifting the box at constant speed?

F (N)

h (m)

mg

1 2 3 4

Constant force acting on object as it is lifted higher and higher.

Page 6: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Gravitational Potential Energy

• What characteristic of our graph is getting larger the higher the object gets lifted?

F (N)

h (m)

mg

1 2 3 4

Page 7: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Gravitational Potential Energy

• What characteristic of our graph is getting larger the higher the object gets lifted?

F (N)

h (m)

mg

1 2 3 4

AREA = Eg

Area = b x h = F x h= mgh

Eg = mgh

Page 8: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Kinetic Energy• Consider a cart being pushed by a force across

a frictionless surface some distance x…

x

F F

Change in Energy will correspond to a force times a distance, therefore

Ek = F x= max ***Recall from studying motion

vf2 = 2ax + vi

2 OR…

ax = ½ vf2 - ½ vi

2

Ek = m(½ vf2 - ½ vi

2)

Ek = ½ mvf2 - ½ mvi

2 So, Ek = ½mv2

Page 9: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Summarizing the Energy Equations

• Elastic Potential Eel = ½ ks2

• Gravitational Potential Eg = mgh

• Kintetic Ek = ½ mv2

Page 10: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Using LOL diagrams to write equations

F

Consider our box once again being lifted by some force external to the system…

Initial Energy Flow Final

Ek Eg Eel Ek Eg Eel

BoxEarth

Working

Ek + Eg + Eel + W = Ek + Eg + Eel

W = mghf

Page 11: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Sample Problem

Energy Flow Diagram

0

EK Eel

Initial Eg

0

Final EK Eg Eel Ediss

A moving cart hits a spring, traveling at 5.0 m/s at the time of contact. At the instant the cart is completely stopped, by how much is the spring compressed? (ignore friction)

Page 12: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

A closer look at Working• Something is working if it transfers energy

into or out of a system… W = E• W is positive when energy is added to a system• W is negative when energy is removed from a

system

Energy Flow Diagram

0

EK Eel

Initial Eg

0

Final EK Eg Eel Ediss

+W - W

Page 13: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Working• In every case, when something is working, a force is

being exerted on an object as the object moves.

• Only forces acting parallel to the motion of the object are Forces acting in a direction parallel to displacement in order for energy transfer to occur.

W = F x

F F

x

F

FN

Fg

f

***In this example, FN and Fg would not change the energy of the system, F would add energy, f woul remove energy

Page 14: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Sample Problem

Energy Flow Diagram

0

EK Eel

Initial Eg

0

Final EK Eg Eel Ediss

A person pushes a 1500 kg car from rest with a force of 500 N. If there was no friction, how fast would the car be moving after pushing for 10 m?

Page 15: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Sample Problem

Energy Flow Diagram

0

EK Eel

Initial Eg

0

Final EK Eg Eel Ediss

A person pushes a 1500 kg car from rest with a force of 500 N. If the coefficient of friction between the car and road is 0.20, how fast would the car be moving after pushing for 10 m?

Page 16: Force on a spring lab… The force applied to a spring and the amount of stretch in the spring are directly proportional… F = ks F This relationship is known

Sample Problem

• A 30 g bullet is fired into ballistic gel. If the gel exerts a force of 50000 N to stop the bullet in a distance of 0.10 m, how fast was the bullet moving? Energy Flow

Diagram

0

EK Eel

InitialEg

0

FinalEK Eg Eel Ediss

Initial Final