# A 35.0-kg crate is initially a

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

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 -Wf

∴ (1/2) m u2 = m g h – (μ FN) 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 – PEspring – Wf

∴ (1/2) m v2 = m g h – (1/2) k x2 – (μFN) (d + x)

∴ (1/2) m v2 = m g {(d + x) sinθ} – (1/2) kx2 – 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)} (0.10m)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) = 36.287Ns

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