# Work 2

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.

Work done = force x displacement in the direction of the force = Fs

Whre s is the distance moved in the direction of the force.
Notice that force and displacement are vectors but work is a scalar.

Work done = Fs]

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.

Work done = F[s cos q]

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.

Example problems
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 30o to the rails. If the frictional force remains the same calculate:
(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

## Work done against the force of gravity

One of the simplest ways doing work is to lift an object up against the pull of gravity. The force on the object would be mg and if it is lifted through a height h the work done (and the gravitational potential energy gained) is mgh.

Work done against the force of gravity = mgDh where Dh is the change in height of the object above some original position measured upwards from that position.

A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB