In the figure, a small block of mass m 0.018 kg can slide along the frictionless

Question

In the figure, a small block of mass m 0.018 kg can slide along the frictionless loop-the-loop, with loop radius R - 11 cm. The block is released from rest at point p, at height h = 3R 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 O and b the top of the loop? r 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? QT

Explanation / Answer

here,

mass of the block , m = 0.018 kg

radius , R = 11 cm = 0.11 m

a)

the work done by gravitational force as it goes from P to Q , W1 = m * g * ( H - R)

W1 = m * g * ( 3 R - R)

W1 = 0.018 * 9.81 * ( 2 * 0.11) J

W1 = 0.0388 J = 3.88 * 10^-2 J

b)

the work done by gravitational force as it goes from P to top of loop , W2 = m * g * ( H - 2R)

W2 = m * g * ( 3 R - 2R)

W2 = 0.018 * 9.81 * ( 0.11) J

W2 = 0.019 J = 1.9 * 10^-2 J

c)

the potential energy when the block is at point P , PEp = m * g * H

PEp = m * g * 3 * R

PEp = 0.018 * 9.81 * 3 * 0.11 J

PEp = 5.83 * 10^-2 J

d)

the potential energy when the block is at point Q , PEq = m * g * R

PEq = 0.018 * 9.81 * 0.11 J

PEq = 1.9 * 10^-2 J

e)

the potential energy when the block is at the top of loop , PEt = m * g * 2 * R

PEt = 0.018 * 9.81 * 2 * 0.11 J

PEt = 3.88 * 10^-2 J

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