Work and Energy

Conservation Law

Work and Energy

Work is a force applied to an object that causes the point of application of the force to move through some distance

Energy is the capacity of an object to do work

Unit: Joules Energy and work are scalar quantities

Work

Is work a dot product or a cross product?

W Fd cos

Example problem

A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F= 50.0 N at an angle of 30.0º with the horizontal. Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3.00m to the right.

Answer

W=Fd cos θ W= 50.0 N (3.00m) cos 30º W = 130 J

Work a Scalar Product

Work is the dot product of Force and Displacement

W F d

Work Done by a Constant Force A particle moving in the xy plane undergoes

a displacement as shown below in a constant force as given below acts on the particle. A) Calculate the magnitude of the displacement and that of the force. B) Calculate the work done on the object.

d i j m

F i j N

( . . )

( . . )

2 0 30

50 2 0

Answerd x y

d

d m

2 2

2 22 0 30

36

( . ) ( . )

.

F F F

F

F N

x y

2 2

2 250 2 0

54

( . ) ( . )

.

W F d

W i j i j

W i i i j j i j j

W J

( . . ) ( . . )

( . . ) ( . . ) ( . . ) ( . . )

50 2 0 2 0 30

50 2 0 50 30 2 0 2 0 2 0 30

10 0 0 6 16

Work Done by a Varying Force Imagine a particle undergoes a very

small displacement Δx The Fx is approximately constant over

this small displacement. Work done can be expressed as

W F xx

Force vs. displacement

01

234

56

1 2 3 4 5 6

Displacement

Fo

rce

Series1

Graph of a Varying Force

Work Done by a Varying Force

W F x

W F x

W F dx

x

xx

x

x

xx

x

i

f

i

f

lim

0

Work

Integration

Work Video

Power

Power is the rate of doing work or of transferring energy.

Pdw

dt

Power Units

Power is measured in joules per second.

1 J/s = Watt 1 hp = 746 W

1 kWh = 3.6 x 106W

Power and Force

If force F causes a particle to undergo a displacement ds, the work done is dW=F•ds.

Since ds/dt is v the power is given by the following formula:

PdW

dt

PF ds

dtP F v

Problem

Consider a car traveling at a steady speed of 60 km/h (16.7 m/s). It encounters a frictional force (rolling and air drag) of 520 N. At what power level does the engine deliver energy to the wheels?

Answer

P = Fv 8.68 kW 11.6 hp