The loaded 150 kg skip is rolling down the incline at 4m/s when a force P is applied to the cable as shown at time t = 0. The force P is increased uniformly with the time until it reaches 600N at t = 4s, after which time it remains constant at this value. Calculate
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Sol.
(i) The skip reverses direction when its velocity becomes zero. We will assume that this condition occurs at t = 4 + Δ t s. The impulse-momentum equation applied consistently in the positive x-direction gives 
(4) (2) (600) + 2 (600) Δ t – 150(9.81) sin 30º (4 + Δ t) = 150 (0 – [–4])
46 Δ t = 1143 Δ t = 2.46s t = 4 + 2.46 = 6.46 Ans.
(ii) Applying the impulse-momentum equation to the entire interval gives 
.
(4) (2) (600) + 4(2) (600) – 150(9.81) sin 30º
(8) = 150 (v – 1 [–4])
150v = 714 v = 4.76 m/s
The same result is obtained by analyzing the interval from t 1 to 8s.
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