A block with mass m = 18.7 kg slides down an inclined plane of slope angle 18.4o

Question

A block with mass m = 18.7 kg slides down an inclined plane of slope angle 18.4o with a constant velocity. It is then projected up the same plane with an initial speed 3.05 m/s. How far up the incline will the block move before coming to rest?

Explanation / Answer

Answer: 1.02 m. Solution: f the mass is sliding with constant velocity =>Friction force Ff = mgsinA => f x mgcosA = mgsinA =>f = tanA = tan 26.3 = 0.49 [ friction factor] On the upward motion the the net force on the mass will be F = f x mgcosA + mgsinA =>F = mg[0.49 x cos26.3 + sin26.3] =>F = mg[0.44 + 0.44] =>F = 0.88mg ; Let the mass travel s meter on the inclined plane =>1/2mv^2 = F x s =>mv^2 = 2 x 0.88mg x s =>s = (4.20)^2/(1.76 x 9.8) = 1.02m

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