A wheel and axle machine having a wheel radius of 25\text{ cm} and an axle radius of 5\tex… — JAMB UTME

A wheel and axle machine having a wheel radius of $25\text{ cm}$ and an axle radius of $5\text{ cm}$ is used to raise a load of mass $80\text{ kg}$ vertically through a height of $10\text{ m}$ . If the efficiency of the machine is $80\%$ , what is the work done against friction during this operation? (Take $g = 10\text{ m s}^{-2}$ )

A
$1,600\text{ J}$
$2,000\text{ J}$ Answer
C
$8,000\text{ J}$
D
$10,000\text{ J}$

Answer

2,000\text{ J}

The useful work done on the load is

W_{\text{out}} = mgh = 80 \times 10 \times 10 = 8000\text{ J}

. Since efficiency is

80\%

, the total work input required is

W_{\text{in}} = \frac{8000}{0.80} = 10,000\text{ J}

. Therefore, the work lost to friction is

W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

Step-by-Step Solution

Calculate the useful work output of the machine

W_{\text{out}} = mgh = 80\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 8,000\text{ J}

Useful work output is the gravitational potential energy gained by raising the load.

Calculate the total work input using the efficiency equation

W_{\text{in}} = \frac{W_{\text{out}}}{\text{Efficiency}} = \frac{8,000\text{ J}}{0.80} = 10,000\text{ J}

Efficiency is defined as

\frac{\text{Work Output}}{\text{Work Input}}

, so Work Input =

\frac{\text{Work Output}}{\text{Efficiency}}

Determine the work done against friction

W_{\text{friction}} = W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

The energy wasted as heat and friction is the difference between total work input and useful work output.

Key Concept

Efficiency of Simple Machines and Work Done Against Friction

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