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Derive an expression for the potential energy of a body near the surface of the Earth. (OR) Calculate the potential energy of the object of mass m at a height h. |
Answer» Solution :(i) The gravitational potential energy (U) at some HEIGHT h is equal to the amount of work required to take the object from ground to that height h with constant velocity.  (ii)Consider a body of mass m being moved from ground to the height h against the gravitational force as shown in Figure. (III)The gravitational force `vec(F)_(g)` acting on the body is, `vec(F)_(g)=-mg hat(j)` (as teh force is in y direction, unit vector is used). Here,negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force `vec(F)_(a)` equal in magnitude but opposite to that of gravitational force `vec(F)_(g)` has to be applied on the body i.e., `vec(F)_(a)-vec(F)_(g)`. This implies that `vec(F)_(a)=-mg hat(j)` (iv)The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains UNCHANGED and thus its KINETIC energy also remains constant. (v) The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h. `U = int vec(F)_(a).d vec(r )=int_(0)^(h)|vec(F)_(a)||d vec(r )|cos theta` (vi)Since the displacement and the applied force are in the same upward direction, the angle between then, `theta = 0^(@)`. Hence, `cos theta^(@)=1` and `|vec(F)_(a)|=mg` and `= |d vec(r )|dr`. `U = ms int_(0)^(h)dr` `U=mg [r]_(0)^(h)=mgh`. |
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