In the figure below, a small block of mass m = 0.025 kg can slide along the fric
ID: 1697508 • Letter: I
Question
In the figure below, a small block of mass m = 0.025 kg can slide along the frictionless loop-the-loop, with loop radius 10 cm. The block is released from rest at point P, at height h = 16R above the bottom of the loop. (For all parts, answer using g for the acceleration due to gravity, and R and m as appropriate.)(a) How much work does the gravitational force do on the block as the block travels from point P to point Q?
(b) How much work does the gravitational force do on the block as the block travels from point P to the top of the loop?
(c) If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is the potential energy when the block is at point P? (d) What is the potential energy when the block is at point Q?
(e) What is the potential energy when the block is at the top of the loop?
(f) If, instead of being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?
remain the same
decrease
increase
Explanation / Answer
(a)given h=19R....work done by gravity from p to q =mgh-mgR=m*g*18*R=0.031*9.8*18*0.15=0.82 J................(b)work done from p to top of loop=mgh-mg*2R=m*g*17R=0.7747J.....(C)when the block is at p potential energy=mgh=0.031*9.8*19*0.15=0.866J....(d)at Q potential energy=mgR=0.031*9.8*.15=0.045J...(e)at top of loop P.E=mg*2R=0.031*9.8*2*.15=0.091J.....(f) remains the same