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A block of mass 10kg is initially at rest on a horizontal rough surface. At a time t = 0, a time dependent force (F) acts on it horizontally as shown in the figure. Force (F) stops acting at t = 7 seconds. Consider that coefficient of static and kinetic friction both are equal to 0.5 and $g = 10 {ms}^{- 2}$ .

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Maximum kinetic energy acquired by the block is 50 J

Maximum kinetic energy acquired by the block is 125 J

Total time for which the block is moving with non zero velocity is 5.4 s

Total time for which the block is moving with non zero velocity is 7.4 s

answer is B.

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Detailed Solution

Block will start moving at F = 25 t

$μ = mg,$

$25 t = 0.5 \times 10 \times 10$

Velocity is maximum $t > 4 \sec$

$\frac{du}{dt} = \frac{25 t - 50}{10}$

$\int_{0}^{v_{\max}} du = \int_{2}^{4} (2.5 t - 5) dt$

$v_{max} = 5 m/s$

$KE = \frac{1}{2} \times 10 \times 25 = 125 J$

For 4s and $< t < 7sec$

$a = \frac{40 - 50}{10} = - 1 {m/sec}^{2}$

$v = 5 - 1 (3) = 2 m/sec$

$a_{2} = - \frac{50}{10} = - 5 {m/sec}^{2}$

$t = \frac{v}{| a^{2} |} = \frac{2}{5} = 0.4 \sec$

Total time $= (4 - 2) + (7 - 4) + 0.4$

$= 2 + 3 + 0.4 = 5.4 \sec$

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