A 11.5-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 11.5-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 5.23 m/s in 0.803 s. In the process, the spring is stretched by 0.201 m. The block is then pulled at a constant speed of 5.23 m/s, during which time the spring is stretched by only 0.0973 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Explanation / Answer

The block's acceleration is
a = 5.23 m/s / 0.803s = 6.51 m/s^2

So the net accelerating force on the block is
fa = m*a = 11.5 * 6.51 N = 74.86 N

The friction force ff also stretches the spring at the other end and is the ONLY stretching force at constant speed:
so ff = 0.0973m * k

(a) So the total stretching force is
fa + ff = 74.86 + 0.0973m * k = k*x = 0.201*k

so k = 74.86 / (0.201 - 0.0973) N/m = 721.89 N/m

(b) ff = k*m*g = 0.0973 * 721.89 N = 70.24N
so the coeff of kinetic friction is

k = 70.24 / (11.5*9.8) = 0.623

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