An elevator cage with its load has a total mass of

PHYSICS · Multi-part · CBSE · Physics · Laws of Motion · Laws of Motion

Problem

500\,\text{kg}

. Starting from rest, it is pulled upward by a cable that exerts a constant upward force of

6000\,\text{N}

for a time interval of

3.0\,\text{s}

. Immediately after this interval the cable force is cut off, and the cage moves under the influence of gravity alone until it momentarily comes to rest at its highest point. Ignore friction in the guiding rails. (a) Calculate the acceleration of the cage during the

3.0\,\text{s}

when the cable force acts. Take

g = 9.8\,\text{m s}^{-2}

. (b) Determine the speed of the cage at the instant the cable force is cut off. (c) Compute the additional height the cage rises after the cable force is removed before coming momentarily to rest. (d) Find the total vertical distance the cage ascends from the instant the motion begins until it reaches its highest point.

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