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JAMB Physics 2023 Paper

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Question : 15

Total: 65

How much net work is required to accelerate a 1200 kg car from 10

\mathrm{m}\,\mathrm{s}^{-1}

\mathrm{m}\,\mathrm{s}^{-1}

1.95× $10^5 j$
1.35× $10^4 j$
7.5× $10^4 j$
6.0× $10^4 j$

Solution:

Concept:

Work done on the car equals its change in kinetic energy.

Explanation:

The initial speed is

v_1 = 10\,\mathrm{m}\,\mathrm{s}^{-1}

.

The final speed is

v_2 = 15\,\mathrm{m}\,\mathrm{s}^{-1}

.

Mass is

m = 1200\,\mathrm{kg}

.

Use the work-energy principle:

W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2

W = \frac{1}{2} m (v_2^2 - v_1^2)

Substitute the values:

W = \frac{1}{2} \times 1200 \times (15^2 - 10^2)

W = 600 \times (225 - 100)

W = 600 \times 125

W = 7.5 \times 10^4\,\mathrm{J}

Answer:

The required net work is

7.5 \times 10^4\,\mathrm{J}

, which is option C.

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