# Work

W = F s cos α

• W   work in J (Joule)  (not to be confused with the weight)
• F     force in N
• s     displacement in m
• α    angle between  F en s

Example 1

See the figure. On an object acts a force F. The object moves 100 m to the right.

The angle α between F and the x-direction is 35 o. The weight is 30 N.

Also work on this object a frictional force (f=5.0 N) and a normal force Fn.

a.Find the work of F

Resolve  F in the x-direction (parallel to the displacement) and in the y-direction (perpendicular to the displacement)

F x = 16.4 N         (Calculation : cos 35 o = Fx/ F         Fx = F cos 35 o)

WF = Fx s = (16.4)( 100) = 1640 J

Also possible :  W = F s cos α = (20)( 100 cos 35) = 1640 J

Remark

If force and displacement have the same direction :  W > 0 J

b. Find the work of the frictional force f

Wf = – f s= – (5.0)(100) = – 500 J

of

Wf = f s cos α =(5.0)( 100 cos 180o) = (5.0)( 100)( -1) = – 500 J

( If force and displacement are oppositely directed then  W < 0 J)

c. Find the work of weight and normal force

W w= W Fn = 0 J

( If F s then : W = 0 J,       α= 90o    cos 90o = 0)

Example 2

An object with a mass of 5.0 kg falls  20.0 m downward.

How much is the work of the weight ?

W = m g = (5.0)( 9.81) = 49.05 N

Ww= W s = (49.05)( 20.0) = 981 J = 9.8 x 102 J

Example 3

we pull a spring ( k = 50 Nm -1) 20 cm.

Calculate the work done by the force excert by the spring.

Method 1

If the spring has been pulled to the right, the force in the spring is directed to the left.

Thus the work of Fspring is  negative .

WFspring = – ½ k x2  = – ½ (50)(0.20)2 = – 1.0 J

Method 2

Determine work with  F versus x graph

F = k x = (50)( 0.20) = 10.0 N 