How do you do this Physics problem?
A 240 kg box initially at rest is being pushed with 5000N by a person for 10 seconds across a floor from point A to point B. The person stops pushing. The frictional coefficient from point A to B is 0.2. At point B the ground becomes frictionless all the way to point C. Between point B and C there is an inelastic collision that occurs with a stationary 50kg crate. At point C the floor becomes really rough and the coefficient of friction is 0.6. The conjoined objects travel through point C until it comes to rest at point D. What is the distance from point C to point D?
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Verified answer
Weight = 240 kg * 9.8 m/s^2 = 2352 N
Friction = Ff = Mu_k * weight (horizontal surface)
Ff = 2352 * 0.2 = 470.4 N
Net Force (Fn) = 500 N - 470.4 N = 29.6 N
F = MA
29.6 = 240 * A
A = 29.6/240 = 0.12 m/s^2
V@B = AT = 0.12 * 10 = 1.2 3m/s
Conservation of Momentum
Momentum before = Momentum after
M1 * V1 = (M1 + M2) * V2
240 * 1.23 = (240 + 50) * V2
296 = 290 * V2
V2 = 1.02 m/s @ point C
KE @ C = 0.5MV^2 = 0.5 * 290 * 1.02^2 = 151.06 J
KE = Work
KE = force * distance
Ff = Mu * W = 0.8 * 2352 N = 1881.6 N
151.06 J = 1881.6 N * d
d = 0.8 m from C