An automobile of mass m = 1100 kg is being towed by a truck whose driver observe

Question

An automobile of mass m = 1100 kg is being towed by a truck whose driver observes that the tension in the rope is T = 3700 N while the rig (truck plus car) is uniformly accelerating such that its velocity increases from vi = 4 m/s to vf = 1.5*vi over the time interval t = 6 s. The wheels turn without slipping on the road and there is some friction within the car.

a) How far does the car travel in this time interval?
b) What is the work done on the car by the rope?
c) What is the work done to overcome friction in the car?

Explanation / Answer

v = u + at a = 2/6 = 1/3 m/s^2 is acceleration distance v^2 = u^2 + 2as 36-16 = 2*1/3 * s s = 30 m distance traveled = 30 m work done by the rope = -3700*30 = -111 kJ frictional force be f so f - T = ma f = T+ma = 3700 + 1100*1/3 = 4066.663 N so work done to overcome friction = 4066.663*30 = 122 kJ

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