Physics 2011 JAMB Past Questions - Question 12: A machine is used to lift a load of 20N through a height of 10m.If the efficiency of the machine is 40%, how much work is done ?

Correct Answer

C

Explanation

Y = Work output/Work input x 100/140 = 20 x 10/x x 100/1 = 500J
Work (W) = Force (F) x Distance (d)

In this case, the force (F) is 20 N (Newtons), and the distance (d) is 10 m (meters). However, since the machine has an efficiency of 40%, you need to account for this reduced efficiency when calculating the work done.

Efficiency (η) is defined as the ratio of useful work output to the total work input. In this case, the useful work output is the work done on the load (lifting it through a height of 10m), and the total work input is the work done by the machine.

Efficiency (η) = (Useful Work Output) / (Total Work Input)

0.40 (40%) = (Useful Work Output) / (Total Work Input)

You want to find the Useful Work Output. To do that, you can rearrange the equation:

Useful Work Output = Efficiency (η) x Total Work Input

Now, you can plug in the efficiency and calculate the total work input:

Useful Work Output = 0.40 x (Force x Distance)
Useful Work Output = 0.40 x (20 N x 10 m)
Useful Work Output = 0.40 x 200 N·m

Useful Work Output = 80 N·m

So, the useful work output is 80 Newton-meters (N·m). This is the work done by the machine, taking into account its 40% efficiency.