Integration of a function can also be done by determining the area under a curve for the specified function.

Dx

Fx

DA

xFA x

dAdxFxFLim xxx

0

dxFdAA x

To find the total area under the curve we sum all the small area segments.

xFAA x

For infinitesimally small area segments:

DA – Small segment of total area defined by rectangle.Fx – height of segment DA.Dx – length of segment DA.

W

The integral gives us the same result as obtained for the summation with more precision. The smaller Dx is the more precisely we can match the curve.

Fx

x

If we have a plot of force vs. position we can obtain an estimate of the amount of work done to move from an initial position to a final position, by finding the total area under the segment of the curve between the two positions.

Work done by a Spring

A spring has a rest length that is defined when the spring is unloaded (no force applied).

The spring resists the applied force by exerting a force in the opposite direction. This force is called a restoring force, since it attempts to return the spring to its rest length (equilibrium).

We can mathematically model this using Hooke’s Law.

xkF

Dx – Spring displacement [m]k – spring constant [N/m]F – Restoring Force [N]

- How far the spring is stretched or compressed from its initial length. All lengths are measured relative to the equilibrium position of the spring (x = 0).

x = 0

xEquilibrium

x = 0

-x

xCompressed

x = 0x

xStretched

F

xkF 0kF

F

xkF

When a force is applied to stretch or compress the spring, what does the spring do?

Opposes applied force

-xx

Fs

x

Fs

-FskxFs

x

xdFW0

cos

0x

sdxFq = 180o, so cosq = -1

x

dxkx0

x

dxkx0

x

xk0

2

2

1

22 02

xk 2

2

1kx

bhA2

1 sFx

2

1

kxx 2

1

2

2

1kx

2

2

1kxAW

2

2

1kxW

Same! The work done by a spring was derived from both the definition of work and the area under the curve. Both methods resulted in the same expression.

The amount of work done to stretch or compress a spring depends on the displacement of the spring from equilibrium. For a pre-stretched (or pre-compressed) spring it is necessary to look at the energy for the initial and final location and then subtract to determine how much the energy changed by.

Let us look at the work required to stretch or compress a spring.

Let us now look at the work done by a net force.

x

x

xdFW0

cos

0

x

x

dxF x

x

x dxma0

x

x

dxdt

dvm

0

x

x

dxdt

dx

dx

dvm

0

q = 0o, so cosq = 1

x

x

dvdt

dxm

0

v

v

vdvm0

v

v

vm0

2

2

1

2

02

2

1

2

1mvmv KE

2

2

1mvKE 2

02

2

1

2

1mvmvKE if KEKEKEW

Maximum kinetic energy that an object can have.

Change in kinetic energy due to a change from an initial velocity v0 to a final velocity v.

Work – kinetic energy theorem.

In order to change the kinetic energy and hence the speed of an object you must do work on the object.

[KE] = J = Nm = kg m2/s2 Energy and work have the same units.

vdt

dx

00 vdt

dx

A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off. The cart comes to rest after traveling 1 m. The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest?

1. 1 m2. 2 m3. 3 m4. 4 m5. 5 m6. impossible to determine

210211 2

1vvmFd

220222 2

1vvmFd

1

2

1021

2

2

2022

22 d

vvm

d

vvmF

1

2

10

2

2

20

d

v

d

v

112

10

2

1012

10

2

202 4

4dd

v

vd

v

vd

2

101 vFd

If vo is doubled, than vo2 is

four times as great, and therefore d2 must be 4 times greater than d1.