A 15.0-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 15.0-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 m/s in 0.530 s. In the process, the spring is stretched by 0.160 m. The block is then pulled at a constant speed of 5.00 m/s, during which time the spring is stretched by only 0.0500 m. (a) Find the spring constant of the spring. N/m (b) Find the coefficient of kinetic friction between the block and the table.

Explanation / Answer

The block's acceleration is
a = 5.00 m/s / 0.530s = 9.433 m/s^2

So the net accelerating force on the block is
fa = m*a = 15 * 9.433 N = 141.51 N

The friction force ff also stretches the spring at the other end and is the ONLY stretching force at constant speed:
so ff = 0.0500m * k

(a) So the total stretching force is
fa + ff = 141.51 + 0.05*k = k*x = 0.160*k

so k = 214.3 / (0.160 - 0.05) N/m = 1948 N/m

(b) ff = k*m*g = 0.05 * 1948 N = 97.41 N
so the coeff of kinetic friction is

k = 97.41 / (15*9.8) = 0.6627

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