physics question

Question

Tommy gives Sally (mass 55.0kg) sitting on a swing an underdog and lifts her 1.25m higher than she was when sitting stationary at the bottom of the swing.

a. What should Sally’s velocity be when she returns to the very bottom position of her swing?

b. When Sally returns to the top of her swing she only reaches a height of 1.11m above the bottom. How much of her mechanical energy was transferred to friction during the swing? Answer: 4.95m/s, Answer: 75.5J

Answer 1

At the highest point of her swing, she will have a potential energy of mgh where m=mass 55 kg, g = 9.8 ms^-2, and h = 1.25 m. This calculates to 674 j.

Upon swinging back to her starting position, the potential energy is converted to kinetic energy

KE = 1/2 x m x v². Equate this to the PE and solve for v. You should get 4.95 m/s.

Upon returning to the top of her swing, her new PE is 55kg x 1.11m x 9.8 ms^-2. The difference between the initial PE and the final PE is the energy lost to friction. This leads to

55kg x 1.25m x 9.8 ms^-2 - 55kg x 1.11m x 9.8 ms^-2 = 75.5 j