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
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
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
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)
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
we pull a spring ( k = 50 Nm -1) 20 cm.
Calculate the work done by the force excert by the spring.
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
Determine work with F versus x graph
F = k x = (50)( 0.20) = 10.0 N
W is equal to the arae under the graph of F versus x
W = ½ (10.0)( 0.20) = – 1,0 J.
W < 0 because the force of the spring and displacement are oppositely directed.