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A solid block of mass 2kg is resting inside a cube as shown in the figure. The cube is moving with a velocity $\vec{v} = | 5 t \hat{i} + 2 \hat{j} | m/s$ . Here, t is time in second. The block is at rest with respect to the cube and coefficient of friction between the surface of cube and block is 0.8. Then : [take $g = 10 {m/s}^{2}$ in negative y direction]

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The total force exerted by the block on the cube is 14N

Force of friction acting on the block is 10N

Force of friction acting on the block is 4N

The total force exerted by the block on the cube is

10 \sqrt{5} N

answer is A, D.

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Detailed Solution

\bar{v} = 5 t \hat{i} + 2 \hat{j}

$a \hat{i} = 5 {m/s}^{2}, a \hat{j} = 0, a_{net} = 5$

Block is rest $F = f$

$f = ma$

$= 2 \times 5 = 10N$

$N = m (g + a j)$

$= 2 (10 + 0) = 20 N$

Total force $R = \sqrt{N^{2} + f^{2}}$

$= \sqrt{{(20)}^{2} + {(10)}^{2}}$

$= \sqrt{400 + 100} = 10 \sqrt{5} N$

Question Image

$f_{\lim} = 0.8 \times 20 = 16$

$F = 10 N < 16$

$∴ f = 10 N$

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