Question

A 17.6-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 3.58 m/s in 1.77 s. In the process, the spring is stretched by 0.250 m. The block is then pulled at a constant speed of 3.58 m/s, during which time the spring is stretched by only 0.0544 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Answer 1

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of

a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ F = n - mg = 0

where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and

• the net horizontal force is

∑ F = F - f = ma

where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have

F - f = kx - µn = kx - µmg = ma

where k is the spring constant and µ is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ F = n - mg = 0

so that n = mg again; and

• the net horizontal force is

∑ F = F - f = 0

This time, F stretches the spring by y = 0.0544 m, so we have

F - f = ky - µmg = 0

Solve the equations in boldface for k and µ :

kx - µmg = ma

ky - µmg = 0

==>   k (x - y) = ma

==>   k = ma / (x - y)

==>   k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

ky - µmg = 0

==>   µ = ky / (mg)

==>   µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574