Work
Work done by a constant force
If the point of application of a force of F Newtons moves through a
distance s metres in the direction of the force then the work done by
the force is given by:
Work = F x s
The unit of
work is the joule (J)
Example 1
The figure shows a box which is pulled at a constant speed across
a horizontal surface by a horizontal rope. When the box has moved a distance of
9m the work done is 54 J. Find the constant resistance to the motion.
Work
= force × distance moved
54
= force × 9
Resistance
to motion = 6N
Work done against gravity
To raise a particle of mass m kg vertically at a constant
speed you need to apply a force of mg
Example 2
Find the work done by a child of mass
16kg whilst climbing to the top of a slide of vertical height 9m.
Work
done against gravity = mgh
=
16 × 9.8 × 9
=
1410J
Work Done Against Friction
Example 3
A horizontal force P pulls a body of
mass 2.75kg a distance of 12m across a rough horizontal surface, coefficient of
friction 0.2. The body moves with a constant velocity and the only resisting
force is that due to friction. Find the work done against friction.
Resolving perpendicular to the plane:
R
= 2.75g
Using F = μR
F
= 0.25 × 9.8 × 2.75
F
= 6.74N
Therefore:
Work
= force × distance moved
=
6.74 × 12
=
80.9J
Exam questions will involve work done against gravity and
friction.
Example 4
A particle of mass 8 kg is pulled at constant speed a distance of
28m up a rough plane which is inclined at 30° to the horizontal. The coefficient
of friction between the particle and the surface is 0.2. Assuming the particle
moves up a line of greatest slope, find:
(a) the work done against friction.
(b) the work done against gravity.
(a) Work done against
friction = force × distance moved
Resolving perpendicular to the plane gives:
R
= 8g cos30º
Using F = μR F = 0.2 × 8g cos30º
F
= 13.58N
Work done against friction = 13.58 × 28
= 380J
(b) Work done against
gravity = mgh
Remember that we need to find the vertical displacement
Work done against gravity = 8 × g × 28 sin30º
=1097.6
Questions 1
1
A man building a wall lifts
75 bricks through a vertical distance of 3.5m. If each brick weighs 5kg, how
much work does the man do against gravity?
2 Find the work done against gravity when a person of mass 90kg
climbs a vertical distance of 32m.
3 A box of mass 12kg is pulled a distance of 25m across a horizontal
surface against resistances totaling 50N. If the body moves with uniform
velocity, find the work done against the resistances.
4 A horizontal force pulls a body of mass 6kg a distance of 11m
across a rough horizontal surface, coefficient of friction 0.25. The body moves
with uniform velocity and the only resisting force is that due to friction.
Find the work done.
5 A horizontal force pulls a body of mass 2.5kg a distance of 25m
across a rough horizontal surface, coefficient of friction 0.3. The body moves
with uniform velocity and the only resisting force is that due to friction.
Find the work done.
6 A smooth surface is inclined at an angle of 30º to the horizontal.
A parcel of mass 18kg lies on the surface and is pulled at a uniform speed a
distance of 7.5m up a line of greatest slope. Find the work done against
gravity.
7 A surface is inclined at an angle 
to the horizontal. A body of mass 78kg lies on the surface
and is pulled at a uniform speed a distance of 7.5m up a line of greatest slope
against resistances totaling 60N. Find:
a) the work done against gravity.
b) the work done against the resistances.
8 A rough surface is inclined at an angle 
to the horizontal. A body of mass 140kg lies on the surface
and is pulled at a uniform speed a distance of 45m up the surface by a force
acting along the line of greatest slope. The coefficient of friction between
the body and the surface is 
Find:
a) the frictional force acting.
b) the work done against friction.
c) the work done against gravity.
9 A rough surface is inclined at an angle 
to the horizontal. A
body of mass 90kg lies on the surface and is pulled at a uniform speed a
distance of 15m up the surface by a force acting along a line of greatest
slope. The coefficient of friction between the body and the surface is 
Find:
a) the work done against friction.
b) the work done against gravity.
Forces at an angle to the direction of motion
Consider a particle P resting on a horizontal surface. If a force
of magnitude F inclined at an angle θ to the horizontal causes the
particle P to move along the surface while remaining in contact with the
surface, you can resolve the force into its horizontal and vertical components.
For a force at an angle to the direction of motion:
Work done = component of force in direction of motion x distance moved in the
same direction.
Example 5
A sledge is pulled across a smooth horizontal floor by a force of magnitude
50 N inclined at 35° to the horizontal. Find the work done by the force in
moving the packing case a distance of 23m.
The horizontal distance
is 23m so we must consider the horizontal component of the force.
Horizontal component = 50 × cos35º = 40.96N
Work done = force × distance moved
= 40.96 × 23 = 942J
Energy
The energy of a body is a measure of the capacity which the body
has to do work. When a force does work on a body it changes the energy of the
body. Energy exists in a number of
forms, but we will consider two main types: kinetic energy and potential
energy.
Kinetic energy
The kinetic energy of a body is the energy
that it possesses by virtue of its motion. When a force acts on a body to
increase its speed, then the work done equates to the increase in kinetic
energy of the body (provided that no other forces are involved).
If a constant force F acts on a body of
mass m, which is initially at rest on a smooth horizontal surface, then after a
distance s the body has velocity v. So
by considering the work equation:
Work done against friction = force × distance
moved
=
F × s
Using the fact that F = ma
and from the constant acceleration
equations
Therefore: work done = m × 
× s
=
Therefore is said to be the kinetic energy of a mass m moving with
velocity v.
Example 6
A particle of mass 0.25kg is moving with a speed of
. Find its kinetic energy.
Using the formula:
:
Example 7
A particle of mass 4 kg is being pulled across a smooth horizontal
surface by a horizontal force. The force does 46 J of work in increasing the
particle's velocity from 
to
. Find the value of p.
The change in Kinetic Energy is given by the formula:
Where u is the initial and v is the final, velocity
respectively. Since there are no other
forces involved the change in Kinetic Energy must equal the work done by the force.
Example 8
A van of mass 1600 kg starts from rest at a set of traffic lights.
After travelling 240 m its speed is 
. Given that the car
is subject to a constant resistance of 450 N find the constant driving force.
Initially the Kinetic Energy is zero.
Finally the
Kinetic Energy is:
Work done against the resistance = force ×
distance moved
=
(F – 450) × 240
The change in
Kinetic Energy equates to the work done. Therefore:
Potential energy
If the particle is raised a distance of h metres the work done
against gravity is mgh joules. The work done against gravity equates to
the increase in potential energy. If the particle is lowered then the potential
energy decreases. When working with potential energy questions it is vital that
a zero potential energy point is decided upon.
Remember: P.E. = mgh
Example 9
A child of mass 14 kg is raised vertically through a distance of
1.8 m. Find the increase in potential energy.
P.E.
= mgh
= 14 × 9.8 × 1.8
=
247J
Example 10
A child of mass 40kg slides 5.5m down a playground slide inclined
at an angle of 
to the horizontal.
Model the child as a particle and the slide as an inclined plane and hence
calculate the potential energy lost by the child.
The important
thing to remember is that potential energy changes after a change in
height. The 5.5m displacement is not the value of h.
means that sin θ = 
.
h
= 5.5 × sin θ
h = 5.5 × = 3.142m
Therefore:
PE
= mgh
= 40 × 9.8 × 3.142
= 1230J
Questions 2
1
A body
of mass 7.5kg, initially moving with velocity 4
, increases its kinetic energy by 55J. Find the final speed
of the body.
2
Find
the increase in kinetic energy when a stationary van of mass 1400kg accelerates
at 4
for 6 seconds.
3