In the figure below, a small block of mass m = 0.023 kg can slide along the fric
ID: 2060800 • Letter: I
Question
In the figure below, a small block of mass m = 0.023 kg can slide along the frictionless loop-the-loop, with loop radius 16 cm. The block is released from rest at point P, at height h = 21R above the bottom of the loop. (For all parts, answer using g for the acceleration due to gravity, and R and m as appropriate.)
http://www.webassign.net/hrw/08_33.gif
(a) How much work does the gravitational force do on the block as the block travels from point P to point Q?
J
(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?
J
(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?
J
(d) What is the potential energy when the block is at point Q?
J
(e) What is the potential energy when the block is at the top of the loop?
J
(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?
increase
decrease
remain the same
NEED EXACT ANSWERS
Explanation / Answer
a) workdone by the gravitational force : change in energy of the block.
at P it has Potential energy = mgh = m*g*21R = 21mgR
at Q : potential energy : m*g*R = mgR
and change in potential energy of the block is the workdone by the gravity : -(mgR - 21mgR) = 20mgR
b) so final point is the top of the loop :
hence potential energy at the loop top : mg*2R = 2mgR
and change in energy : -(2mgR - 21mgR) = 19mgR : workdone by gravity
c) at point P : mgh = mg*21R = 21mgR
d) at point Q : mgh = mg*R = mgR
e) at loop top : mg*2R = 2mgR
f) remains same as the workdone by the gravity depends on the variation of the height of the block and the same reason accounts for the blocks potential energy.