If an object of mass 2 kg is thrown up from the ground reaches a height of 5 m and falls back to the Earth (neglect the air resistance) Calculate
(a) The work done by gravity when the object reaches 5 m height
(b) The work done by gravity when the object comes back to Earth
(c) Total work done by gravity both in upward and downward motion and mention the physical significance of the result

Explanation

When the object goes up, the displacement points in the upwand direction whereas the gravitational force acting on the object points in downward direction. Therefore, the angle between gravitational force and displacement of the object is `180^(@)`
(a) The work done by gravitational force in the upward motion Given that `dr = 5m and F = mg`
`W_(up)=Fdrcos theta = mgdrcos 180^(@)`
`W_(up)= 2xx10xx5xx(-1)=-100 joule` `[cos180^(@)=-1]`
(b) When the object falls back, both the gravitational force and displacement of the object are in the same direction. This implies that the angle between gravitational force and displacement of the object is `0^(@)`
`W_(down)=Fdrcos0^(@)`
`W=_(down)=2xx10xx5xx(1)=100joule` `[cos0^(@)]`
(c) The total work done by gravity in the entire trip (upward and downward motion) `W_("total")= W_("up") w_("down")=-100" joule + 100 joule "= 0`
It implies that the gravity does not transfer any energy to the object. When the object is thrown upwards, the energy is transferred to the object by the extemal agcocy, which means that the object gains some energy. As soon as it comes back and hits the Earth, the energy gained by the object is transferred to the surface of the Earth (i e., dissipated to the Earth).