A 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon’s (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be ?different? if there were appreciable air resistance?

Solution 18E Here, we shall have to use the equations of potential energy and kinetic energy. The mathematical expression for potential energy is P = mgh…..(1), where m is the mass of the body, g is the acceleration due to gravity and h is the height. 1 2 The mathematical expression for kinetic energy is K = m2 …..(2), here v is the speed of the body. Given, initial speed of the watermelon u = 0 Its mass m = 4.80 kg Height of the building h = 18.0 m 2 We know, acceleration due to gravity g = 9.80 m/s The watermelon is displaced by 18.0 m when it falls from the roof of the building to the ground. (a) Therefore, work done by gravity on the watermelon W g = mgh 2 W g = 4.80 kg × 9.80 m/s × 18.0 m W g = 846.7 J W g 847 J Therefore, the approximate...