A 5.00 kg block rests on a 10.0 kg block as shown in the diagram. The lower mass

Question

A 5.00 kg block rests on a 10.0 kg block as shown in the diagram. The lower mass rests on a
frictionless surface, but the coefficient of static friction between the two blocks is s. A horizontal, variable force, F, acts
directly on the LOWER block, causing both blocks to accelerate from rest without slipping relative to each other. It is
found by experiment that the position of the two blocks varies as x =5t^3/27
It is found from experiment that the static
friction between the blocks reaches its limiting value (maximum possible value) at t = 3.00 seconds. Calculate the
coefficient of static friction, s, between the two

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Explanation / Answer

Given Mass of the upper block is , m1 = 5.0 kg Mass of the lower block is , m2 = 10.0 kg The positionof the two blocks is , x = 5t3/7 Net force on the upper block is m1a = sm1g    ----1 m1a = sm1g    ----1 Net force on the lower block is m2 a = F - sm1g ---2 Here  sm1g    =  sm2g but oppsite in direction     whereas acceleration a = d2x / dt2                                    v = dx/dt = d/dt(5t3/7) = 15t2/7                                     a = 30 t /7 The static friction reaches its value ta t = 3.0 s                                      a = 30(3.0s ) /7 = 12.85 m/s^2 Taking eq 1,   a =  sg                        s = 12.85 m/s^2 / 9.8 m/s^2                         s = 1.3 Thus, the static friction between the two blocks is 1.3                                    v = dx/dt = d/dt(5t3/7) = 15t2/7                                     a = 30 t /7 The static friction reaches its value ta t = 3.0 s                                      a = 30(3.0s ) /7 = 12.85 m/s^2 Taking eq 1,   a =  sg                        s = 12.85 m/s^2 / 9.8 m/s^2                         s = 1.3 Thus, the static friction between the two blocks is 1.3

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