Work is said to be done when a force moves its point of application a distance in the direction of the force. Work is a scaler quantity. The workdone on a body can be defined as the product of force applied to move the body and distance in the direction of the force. It is measured in joules (J).
$$ \text{Workdone, W} = F × s $$ $$\text{where, s is in metres} $$ $$ \text{F is in Newton} $$ $$ \text{But, F = mg} $$ $$ w = mg × s $$A reasonable amount of work is needed to lift a body upwards. When a ball is thrown vertically upwards on application of a force, workdone against gravity is said to be done on that ball. Similarly, when a man moves up a set of steps, he is working against gravity. The workdone against gravity on an object lifted upwards at a height, h is given by: $$ w = mgh $$ Also, when a body of mass m, falls from a height h, under the effect of gravity, the workdone towards gravity is given by: $$ w = mgh $$
If a body of mass m, rolls down or up a hill of height, h along a plane inclined at an angle θ, the workdone on
the body is dependent on the direction moved by the object.
If the body moves to the horizontal direction,
$$ w = Mgcos\theta × s $$
If the body moves to the vertical direction,
$$ w = mgsin\theta × s $$
When a body is rolled down a hill of height, h and length l, the workdone is the product of the force applied and
the height of the hill and not the length of the slope.
$$ w = mgh $$
Example 1: A bead travelling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.001kg and the coefficient of friction between the bead and the wire is 0.1, determine the workdone by friction. (g = 10m/s²) (JAMB)
Example 2: The efficiency of a machine is 80%. Calculate the workdone by a person using the machine to raise a load of 300kg through a height of 4m. (g = 10m/s²) (WAEC)
The efficiency of a machine is given by:
$$ Eff= \frac{\text{work output}}{\text{work input}} × 100 $$
$$ Eff = 80% $$
$$ m = 300kg $$
$$ h = 4m $$
$$ \text{work input} = \text{workdone by person} $$
$$ {W_i} = ? $$
$$\text{workdone by machine} = \text{work output}$$
$$ {W_o} = mgh $$
$$ {W_o} = 300 × 10 × 4 = 12000J $$
$$ Eff = \frac{W_o}{W_i} × 100 $$
$$ 80 = \frac{12000}{W_i} × 100 $$
$$ {W_i} = \frac{12000 × 100} {80} $$
$$\text{Workdone by person} = 15000J $$
Example 3: A boy drags a bag of rice along a smooth horizontal floor with a force of 2N applied at an angle of 60⁰ to the floor. The workdone after a distance of 3m is? (JAMB)
Example 4: A constant force of 40N acting on a body initially at rest gives an acceleration of 0.1m/s² for 4s. Calculate the workdone by the force. (JAMB)