Question

+ -/6 points HRW6 8.P.006. In Fig. 8-28, a small block of mass m 2.0 kg can slide along the frictionless loop-the-loop. The block is released from rest at point P, at height h = 10R above the bottom of the loop. Lon Figure 8-28 (a) How much work does the weight of the block do on the block as the block travels from point P to point ? GR (b) How much work does the weight of the block do on the block as the block travels from point to the top of the loop? OR Assume that the gravitational potential energy of the block Earth system is taken to be zero at the bottom of the loop. (c) What is the potential energy when the block is at point P? GR (d) What is the potential energy when the block is at point Q? GR (e) What is the potential energy when the block is at the top of the loop? (1) 11, instead of being released the block is given some initial speed downward along the track, do the answers to (a) through (o) increase, decrease, or remain the same? remain the same increase decrease

Answer 1

Here we apply concept of conservative force and definition of gravitational potential energy and work done by conservative force.

h=lor m = 2.0kg As gravity is conservative force to work done by gravity, Wg -- (Uf-U;) U = final Potential energy i = Paitial potential energy Let bottom pont of loop as zero potential > J; = mg t = lomo R m=279 Vi= 209R y = mg R = QgR. =Wg so = Ui-Uf = 20gr-2 GR Wglese = 18 gr A from P to top of loop Here U = 209R Hy = 2mg R = tgR to Wp- too = Ui - Uf = 209R-4gR Wot too = 16 GR 2. At bottom of loop, potential energy wh zero then at ip', Op = mgh =20gR

- e At point Q Wo = ng R = 2gR) at top of loop, Utop - mg (22) Utop= 4gR as m=27g. A Mo, as seen above, Work done by gravity de independent of initial speed of object se remains the same.