Question

A 35.0-kg crate is initially at rest at the top of a ramp that is inclined at an angle θ = 30◦ above the horizontal. You release the crate and it slides 1.25 m down the ramp before it hits a spring attached to the bottom of the ramp. The coefficient of kinetic friction between the crate and the ramp is 0.500 and the constant of the spring is k = 6000 N/m. What is the net impulse exerted on the crate from the instant it hits the spring until it has compressed the spring by 10.0 cm?

Answer 1

Solution :

Given :

m = 35 kg

θ = 30^◦

d = 1.25 m

μ = 0.5

k = 6000 N/m

x = 10 cm = 0.10 m

.

According to the conservation of energy :

Kinetic energy of the block when it hits the spring = ΔPE - W_f

∴ (1/2) m u² = m g h - (μ F_N) d

∴ (1/2) m u² = m g (d sinθ) - μ (mg cosθ) d

∴ u² = 2g (d sinθ) - 2μ (g cosθ) d

∴ u² = 2 g d (sinθ - μ cosθ)

∴ u² = 2(9.81 m/s²)(1.25 m) { sin(30) - (0.5)cos(30)}

∴ u² = 1.643 m²/s²

∴ u = 1.282 m/s

.

Similarly :

Kinetic energy of the block when it has compressed the spring = ΔPE - PE_spring - W_f

∴ (1/2) m v² = m g h - (1/2) k x² - (μ F_N) (d + x)

∴ (1/2) m v² = m g {(d + x) sinθ} - (1/2) k x² - μ (mg cosθ) (d + x)

∴ v² = m g {(d + x) sinθ} - (k/m) x² - μ (mg cosθ) (d + x)

∴ v² = 2 g (d + x) (sinθ - μ cosθ) - (k/m) x²

∴ v² = 2(9.81 m/s²)(1.25 m + 0.10 m) { sin(30) - (0.5)cos(30)} - {(6000 N/m) / (35 kg)} (0.10 m)²

∴ v² = 0.06 m²/s²

∴ v = 0.245 m/s

.

Since, Impulse exerted on the crate = change in momentum

Therefore : Impulse exerted on the crate = mu - mv = m (u - v) = (35 kg)(1.282 m/s - 0.245 m/s) = 36.287 Ns