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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» <html><body><p></p>Solution :(i) The gravitational potential energy (U) at some <a href="https://interviewquestions.tuteehub.com/tag/height-1017806" style="font-weight:bold;" target="_blank" title="Click to know more about HEIGHT">HEIGHT</a> h is equal to the amount of work required to take the object from ground to that height h with constant velocity. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V01_C04_E02_105_S01.png" width="80%"/><br/>(ii)Consider a body of mass m being moved from ground to the height h against the gravitational force as shown in Figure. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>)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)` <br/>(iv)The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains <a href="https://interviewquestions.tuteehub.com/tag/unchanged-2316761" style="font-weight:bold;" target="_blank" title="Click to know more about UNCHANGED">UNCHANGED</a> and thus its <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy also remains constant. <br/> (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. <br/> `U = int vec(F)_(a).d vec(r )=int_(0)^(h)|vec(F)_(a)||d vec(r )|cos theta` <br/> (vi)Since the displacement and the applied force are in the same upward direction, the angle between then, `theta = 0^(@)`.<br/>Hence, `cos theta^(@)=1` and `|vec(F)_(a)|=mg` and <br/> `= |d vec(r )|dr`. <br/> `U = ms int_(0)^(h)dr` <br/> `U=mg [r]_(0)^(h)=mgh`.</body></html>


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