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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:

m=1200kg, $V_1$ = $10\,\mathrm{m}\,\mathrm{s}^{-1}$ $V_2$ = $15\,\mathrm{m}\,\mathrm{s}^{-1}$ , w= ?

work=►K.E = $K.E_2$ = $K.E_2$ - $K.E_1$

⇒work=

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

⇒work=

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

)

⇒work=

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

⇒work = 600 × (225 -100)

⇒work= 600 × 125

⇒work= 7.5×

10^4 j

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