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(i) The gravitational poetential 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 cosntant velocity. <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> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V02_G_MQP_01_E03_008_S01.png" width="80%"> <br> (iii) The gravitational force `vecF_(g)` acting on the body is, `vecF_(g) = -mg hatj` (as the 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 `vecF_(a)` equal in magnitude but opposite to that of gravitational force `vecF_(g)` has to be applied on the body i.e., `vecF_(a)- vecF_(g)`. This implies that `vecF_(a) = -mg hatj`. <br> (iv) The positive sign implies that the applied force is in vertically upward direction. Here, when the body id lifted up its velocity remains unchanged and thus its kinetic 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 form the ground to that height h. <br> `U= int vecF_(a).d vecr= int_(0)^(h)|vecF_(a)||d vec r| cos theta` <br> (vi) Since the displacement and the applied force are in the same upward direction, the angle between them, `theta = 0^(@)`. <br> Hence, `cos theta^(@) = 1 " and "|vecF_(a)|= mg` and <br> `=|d vecr| dr.` <br> `U= mg int_(0)^(h) dr` <br> `U= mg[r]_(0)^(h) = mgh`.