In the figure, a small block of mass m = 0.049 kg can slide along the frictionle

Question

In the figure, a small block of mass m = 0.049 kg can slide along the frictionless loop-the-loop, with loop radius R = 18 cm. The block is released from rest at point P, at height h = 4R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop?

Explanation / Answer

use energy equation
PE = mgh

remember to use meters in your calculations

point P is 72 cm or 0.72 m above the bottom of the 18 cm or 0.18 m radius loop

if q is the height above the bottom of the loop for the point Q

a) the height difference between P and Q is (0.72 - q)
so the work is mgh
E = 0.049(9.81)(0.72 - q) for point Q

b) E = 0.049(9.81)(0.72 - 0.4) = 0.15 J for the top of the loop

c) PE = 0.049(9.81)(0.72) = 0.34 J
d) PE = 0.049(9.81)(q)
e) PE = 0.049(9.81)(0.4) = 0.192 J

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.