A stone is thrown from a height (h_o) above a level field, leaving the hand at a

Question

A stone is thrown from a height (h_o) above a level field, leaving the hand at a 40 degree angle. Ignore all effects of air resistance.

a) Compute the work done by gravity as the stone follows its trajectory back to the initial height. Recall that the motion can be divided into motion in the vertical direction and motion in the horizontal direction.

b) Show, by applying the work-energy theorem, that the speed of the stone when it reaches (h_o(initial)) again is identical to the speed it had when it left the hand.

Explanation / Answer

Work done by force = Force x displacement in the direction of force.
Force of gravity is in the vertical dircetion. To find work done by gravity we need to find displacement of body in vertical direction.
As stone comes back to initial height, it's displacement in the vertical direction = 0
Hence work done by gravity = 0

b) Work done by forces = change in kinetic energy of object
Only force acting on object is fore of gravity. As work done by gravity is zero, by the time it reaches it's initial height, change in it's kinetic energy = 0
Kinetic energy = 1/2 m v^2, mass of stone is constant, so there is no change in v. Hence speed of the stone when it reaches initial height is same as, when it left hand

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