I can\'t use kinematic to solve it A 5 meter wide, 200 kg block is travelling al

Question

I can't use kinematic 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? Friendly Formulae 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 sigma F = m a_net

Explanation / Answer

length, L = 5 m

initial velocity , u = 20 m/s

m = 3 Kg

M = 200 Kg

let the coefficient of friction is u

a)

frictional force , f = u * m * g

Using work energy theorum

change in kinetic energy = work done by friction

0.5 * m * (0 - 20^2) = - u * m * 9.8 * 5

solving for u

u = 4.08

the coefficient of kinetic friction must be 4.08

b)

work done to stop the 200 Kg block = change in kinetic energy

work done to stop the 200 Kg block = - 0.5 * M * u^2

work done to stop the 200 Kg block = - 0.5 * 200 * 20^2

work done to stop the 200 Kg block = -40000 J

the work done to stop the 200 Kg block is -40000 J

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