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
