Mechanics Tutorial 13 - Work, Energy, and Power

 Contents
##### Work

Work is defined as:

The product of force and the distance moved (1 mark) in the direction of the force (1 mark).

Work = Force × distance moved in the direction of the force.

If the applied force is in the same direction as the movement, we see:

This uses the formula familiar from GCSE:

If the applied force is at an angle the movement, we see:

This time the formula becomes:

Note that:

• Units are newton metres (Nm) or joules (J).

• Work is actually a scalar quantity despite being the product of a vector quantity.

• Normally we consider the line of action of the force and the line of displacement to be at zero degrees to each other.  The cosine of zero is 1.  However if we have the line of action of the force and the displacement at an angle, we have to use the cosine function to take this into account.

• When work is done, there must be movement.  This can result in acceleration, a rise in temperature, or deformation in shape.

 It is wrong to say that Work = Force × displacement.  If we push the box in the animation (see link above) back to where it started, the displacement is 0, but the distance in the direction of the force is 10 metres.

We can plot a graph of force against distance moved:

In this graph we see that a small force (F1) is applied over a longer distance (s2) and a larger force (F2) is applied over a shorter distance (s1).  The areas under both graphs, the work done, is the same, W.

If the force varies as in the graph below, the work done is still the area.

Be careful here.  The start point is not zero, but s1,  The work done = area of the triangle.  Therefore:

If we have an irregular variation of the force, the work done is still the area under the graph.

In this case we have to count the squares.

 Question 1 A car owner is trying to bump-start his car, but he cannot get it to move.  Sweat is pouring off him.  Explain why he has done no work. Question 2 A horse is pulling a barge along a canal as shown in the diagram.  It pulls the barge with a force of 1000 N a distance of 75 m.  The angle the rope is at 15o to the direction of travel.   The situation is shown in the diagram:   (a) Explain why the answer is NOT 75000 J? (b) What is the work done by the horse?

Energy and work are very closely related.

• Energy is the ability to do work.  When work is done, energy is transferred.

• Energy comes in many forms.

• Some kinds of energy can be stored, while others cannot.

• Energy is always conserved.

 Question 3 A box is pushed 5 m across a room with a force of 30 N.  What is the work done and how much energy is used?

Power is the rate at which energy is used or the rate at which work is done.

Power = energy transferred (J) = work done (J)

time taken (s)             time taken (s)

Units of power are watt (W).

• 1 watt = 1 joule per second.

• Also kilowatt (kW).  1kW = 1000 W;

• megawatt (MW). 1 MW = 1 × 106 W.

 It takes 20 seconds to push the box in Question 3 across the room.  What is the power?

We can also relate power, force and speed:

• Work done = force x distance moved.

W = Fs

• Power = energy ÷ time.

• Speed = distance ÷ time:

• So we can write:

P = W/t

• Therefore:

P = Fv

Power (W) = force (N) × speed (m s-1)

 An electric locomotive is pulling a train at a constant speed of 30 m s-1.  The train has a rolling resistance of 100 kN.    (a) What force must the locomotive produce?  Explain your answer. (b) What power does it use?

Conversion of Work to Energy (Welsh Board)

When work is done, energy is transferred.  Energy is needed for us to do a job of work.  Assuming no losses, we can say:

Work done = Energy transferred

Consider a toy truck of mass m at rest on a frictionless track:

A constant force F is applied and the truck moves through a distance, s.  (Note, for consistency, we are using s for distance, rather than x.) We know that:

W = Fs

We know from Newton II that:

F = ma

So we can easily see that:

W = mas

We know also from our equations of motion that:

v2 = u2 + 2as

We can ignore the u2 term in this argument as the truck is at rest, so u = 0.  We can multiply the equation by m on both sides:

mv2 = 2mas

Since W = mas, we can say:

mv2 = 2W

Therefore:

 What is the significance of this?

Now suppose the truck was travelling at speed u before the force was applied.

We know from our equations of motion that:

v2 = u2 + 2as

We can multiply the equation by m on both sides:

mv2 = mu2 + 2mas

Since W = mas, we can say:

mv2 = mu2 + 2W

So we can rearrange this to:

Since W = Fs, we can also write:

 A train of mass 1.50 × 105 kg is travelling at a speed of 20 m s-1.  A constant force of 100 kN is applied to the train over a distance of 500 m.  What is the new speed at which the train is travelling?