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 F_{n}.

**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 }= F_{x}/ F F_{x} = F cos 35^{ o})

W_{F} = F_{x}_{ }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

W_{f }= – f s= – (5.0)(100) = – 500 J

of

W_{f}_{ }= f s cos α =(5.0)( 100 cos 180^{o}) = (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= 90^{o} cos 90^{o} = 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

W_{w}= W s = (49.05)( 20.0) = 981 J = 9.8 x 10^{2} 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 F_{spring}_{ }is_{ }negative .

W_{Fspring}_{ }= – ½ k x^{2} = – 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

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.