Can\'t use kinematic equations to solve it A 5 meter wide, 200 kg block is trave

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Can't use kinematic equations to solve it

A 5 meter wide, 200 kg block is travelling along a frictionless surface with a velocity of 20 m/s. Sitting atop it, at the leftmost edge, is a much smaller, 3kg block. Refer to Figure 2. a. If the 5 meter wide block is suddenly and instantaneously brought to a complete stop, what must the coefficient of kinetic friction between the two blocks be for the much smaller block to come to a stop at the rightmost edge of the 5 meter block? b. How much work had to be done to stop the 200 kg block? K.E. = 1/2mv^2 P.E._grav = mgh P.E._spring = 1/2 k (Deltax)^2 E_Total = K.E + P.E. E_T,i = E_T,f W_n.c. = DeltaE_T weight Force: w = , where g = 9.81 m/s^2 Friction Force Magnitude: F_friction = mu_k N, where N is the Normal Force Magnitude

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