When a force moves an object work is done on the object and energy is
converted from one form to another.

The units for work are Joules (J).

The greater
the force and the further it moves the greater the work done. Work is the product of the force
and the distance moved by the object in the direction of the force. Strictly we should use the
term displacement here instead of distance.

Whre s is the distance moved in the direction of the force.

Notice that force and displacement are vectors but work is a scalar.

The diagram in Figure 1 shows a truck being pulled at a constant velocity a distance s along a pair of rails by a force F. If there were no friction between the rails and the truck the force needed to keep the truck moving would be zero. However if there is a force of friction F' between the rails and the truck once the truck is moving it will require a force F (= F') acting left to right to keep the truck moving at a constant velocity. The work done on the truck is Fs.

In Figure 2 the force is
applied at an angle q to the rails. The truck has a displacement s along the rails and the truck
moves a distance s cosq in the direction of the force F.

The work done on the truck
is Fs[cos q]. This is often thought of as [Fcosq]s but more correctly it should be
thought of as F[scosq] as shown in the diagram. Mathematically they both come to the same
answer.

In this case Fcosq = F' and so F will need to be large to counteract the frictional
force.

In the extreme case with the force acting at right angles to the rails the truck
would not move along the rails and the work done on the truck would be zero. Clearly the
most effective direction in which to apply the force to the truck if we want it to move along the
rails is parallel to the rails as shown in Figure 1.

The work done on an object is the area
under the line in a force-displacement graph. This applies if the force is constant (Figure
3(a)) or varying (Figure 3(b)).

See the file about muscles and energy in the 14-16/Mechanics section of the site to understand why holding an object at rest in your hand requires energy to be transformed and therefore work to be done.

1. Calculate the minimum work done to pull a truck 8 m at constant velocity along a pair of rails if the frictional force opposing the motion is 100 N

In this case the minimum work done would be when the force is parallel to the rails.

Work done = Force x displacement = 100x8 = 800 J

2. The force is now applied at an angle of 30

(a) the force required to keep the truck moving at the same constant velocity along the rails

(b) the work done in moving the truck 8 m along the rails

Force (F) cos q = frictional force = 100 N

Therefore F = 100/cos 30 = 115.5 N

Work done = 115.5x8 = 923.8 J