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# Work

Module by: Rod Macdonald. E-mail the author

## WORK

Definition 1: Work
is defined as the product of the magnitude of a force applied and the magnitude of the displacement in the direction of the force.

W=F×sW=F×sW = F times s

w = work in Joules (J)
F = force in Newtons (N)
s = displacement in metres (m)

### Note:

Work is a scalar

Definition 2: Joule
One joule of work is done when a force of one Newton acts through a displacement of one metre in the direction of the force. 1J = 1N.m

### Example 1: Work Done if the Force and Displacement are in the Same Direction.

W=Fs W=(20N)(5m) W=100Nm WFs W(20N)(5m) W100Nm
(1)

W=Fs WFs
(2)

### Example 2: Work Done if the Force and Displacement are not in the Same Direction.

Fx is the force in the direction of the displacement.

W = F x s F x = F x cos θ F x = ( Fcos θ ) s W = F x s F x = F x cos θ F x = ( Fcos θ ) s "W" = "F"_"x" cdot "s" newline "F"_"x" = "F" "x" cdot cos%theta newline "F"_"x" = (Fcos%theta) s

### Example 3

Fx is the force in the direction of the displacement.

W = F x s F x = F x cos θ F x = ( Fcos θ ) s W = F x s F x = F x cos θ F x = ( Fcos θ ) s "W" = "F"_"x" cdot "s" newline "F"_"x" = "F" "x" cdot cos%theta newline "F"_"x" = (Fcos%theta) s

Work Done if the Force and the Displacement are Perpendicular.

### Example 4

If the force is perpendicular to the direction of the displacement there is a zero component of the force in the direction of the displacement.

W=FcosθqsW=FCos90°sW=F(0)(s)W=0W=FcosθqsW=FCos90°sW=F(0)(s)W=0"W"="F" cdot cos%theta q cdot s newline "W" ="F" "Cos"90^circ cdot s NEWLINE "W" = F(0) CDOT (s) NEWLINE "W" = 0

(Thus no work is done!)

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