Question

1. A 9.6 kg particle is attached to a rope of length L. 1.7 m. It is released from the horizontal position A, and it swings as a pendulum. Once the particle reaches point B (a distance h 1.2 m below), you push horizontally on the particle with a varying force F such that it slows down and comes to rest at the lowest point C a. Calculate the speed of the particle at point B. b. Calculate the total work done on the particle as it falls from point A to c. Calculate the work done by gravity on the particle as it falls from point A d. Calculate the work done by the rope on the particle as it falls from point A e. Calculate the work done by the force F on the particle as it falls from point C. to point C. to point C. point A to point C.

1

Answer 1

mgh = 1/2 mv^2 v = squareroot 2gh = squareroot 2(9.8)(1.2) = 4.84 m/sec Work Done by Gravity to move from A to B W_g = mgh = (9.6)(9.8)(1.2) = 112.896 Joule Work Done by force(F) to speed down form B to C W_F = -mgh� h� = L - H = -(9.6)(9.8)(1.7 - 1.2) = -47.04 Joule Total Work Done = (112.896) - (47.04) = 65.85 Joule Work Done by gravity = W_g = 112.896 J Work Done by Rope rightdoublearrow FS cos theta