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Work Energy and Power

Section: Physics

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Question : 13 of 26

Marks: +1, -0

The linear momentum of a body of mass 8 kg is

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

. If a constant force of 24 N acts on the body in the direction of motion of the body for a time of 3 s , then the increase in the kinetic energy of the body is

[AP EAPCET 21 May 2025 S1]

$480\,\mathrm{J}$
$540\,\mathrm{J}$
$270\,\mathrm{J}$
$240\,\mathrm{J}$

Solution:

\;p=mv=24\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}

\; \therefore u=\frac{24}{m}=\frac{24}{8}=3\,\mathrm{m}/\mathrm{s}

\;\;\text{ Impulse, }\; J=F \cdot \Delta t

\;\;\;\;\;\;\; =24 \times 3=72\,\mathrm{Ns}

\therefore \;\; J=\Delta p=72\,\mathrm{Ns}

\therefore

Final momentum

p_f\;=p_i+\Delta p=24+72

\;=96\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}

\;\;\therefore\;\; v_f\;=\;\frac{p_f}{m}=\;\frac{96}{8}=12\,\mathrm{m}/\mathrm{s}

Thus, increase in kinetic energy,

\Delta K\;=K_f-K_i

\;=\;\frac{1}{2} m v_f^2-\;\frac{1}{2} m u^2

\;=\;\frac{1}{2} m (v_f^2-u^2)

\;=\;\frac{1}{2} \times 8 (12^2-3^2)

\;=4(135)=540\,\mathrm{J}

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