1. A 2.00 kg ball is attached to the ceiling by a 1.00 m long string. The height
ID: 1693066 • Letter: 1
Question
1. A 2.00 kg ball is attached to the ceiling by a 1.00 m long string. The height of the room is 3.0 m. (a) What is the gravitational potential energy of the ball relative to the ceiling? (b) relative to the floor? (c) relative to a point at the same elevation as the ball?Explanation / Answer
Gravitational potential energy due to attraction of earth is given by the equation (PE) = (GMm) / (R+h) G=Universal gravitational costant;M=Mass of the earth;m=mass of the ball ;R=radius of earth ;h = Height of the ball from the suface of the earth. G=6.67*10^-11; M=7.4*10^22Kg R= 1.7*10^6m h=3-1=2m; m=2.00Kg ==>Gravitational potential energy (PE) = (6.67*10^ -11 * 7.4*10^22 * 2)/1.7*10^6 (his very small it is neglected). (PE) = (98.716 *10^11)/1.7 *10^6 (PE) = 58.07 *10^5 j In all the three cases the (PE) will be same,because the gravitational PE is calculated with respect to it's height from surface of earth. This h value is same in all the three cases,hence PE also will be same.
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