Question

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Answer 1

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Given :-

Mass of the box = m = 20 Kg

Distance = 4 m

Coefficient of Friction = μ = 0.15

Acceleration due to gravity = g = 10 ms-²

Acceleration = 2 ms-²

${a}_{max}amax = μg$

${a}_{max}amax = 1.5 ms-²$

Now,

Relative Acceleration =${a}_{max}amax - a$

${a}_{r} = 1.5 - 2ar=1.5−2$

${a}_{r} = 0.5 ms-²ar=0.5ms−²$

In the opposite side.

t = √2S/ar

t = √2×4/0.5

t = 4 s.

S' = 1/2 ${a}_{r}ar t²$

S' = 1/2 × 0.5 × 4 × 4

S' = 4 m

Answer 2

Answer:

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