At the instant shown, the crate A of weight w_A is sliding down a rough incline

Question

At the instant shown, the crate A of weight w_A is sliding down a rough incline whose coefficient of kinetic friction mu_k = 0.25 with a speed v_A = ft/s. ft strikes a stationary crate B of weight w_B that is attached to an unstretched compression spring at the bottom of the incline. The two crates stick together and slide a further distance d_s before coming to rest. Model the crates as particles and take the spring constant k = 300 Ib/ft, w_A = 10 lb, w_B = 2.5 lb, d = 2 ft, v_A = 20 ft/s, g = 32.2 ft/s^2, and theta = 30 degree. Determine the preimpact velocity of the crate A. Determine the postimpact velocity of the crate A and crate B. Note that the imp; is plastic (i.e., ..coefficient of restitution e = 0 and the two crates stick together). Determine the distance d_s the spring will compress for the system (crates A an stuck together) to come to rest. Determine the average normal force exerted on crate A during the impact if the time of impact between the box and block is 0.0015 s.

Explanation / Answer

accleration of block A = gsi n30 - 0.25 ( g) cos 30 = 2.778 m/s^2 = 9.114173 ft/s^2

a) Va ( before impact) = sqroot ( 20^2 + 2x 9.114x 2) = 20.89 ft/sec

b) conserving momnetum

( 10 lb ) ( 20.89 ) = ( 10 + 2.5) V

v = 16.71 ft /sec

c) F = kx ( Hooke's law)

12.5 = 300 (x)

x= 0.04167 ft apprx

d) force = change in momentum/time

mass of A = 4.535 kg

Force = 4.535 ( 20.89 - 16.71) ft/sec / 0.0015 sec= 4.535 x 1.382268 m/sec / 0.0015 = 4179.056 N apprx

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