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In Fig 8-33, a small block of mass m=0.032kgcan slide along the frictionless loop-the-loop, with loop radius R=12cm. The block is released from rest at point P, at height h=5.0R above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point Pto a) point Q(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 (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward alongthe track, do the answers to (a) through (e) increase, decrease, or remain the same?

Short Answer

Expert verified

a) The work from point Pto point Q is 0.15J

b) The work from point Pto top of the loop is 0.11 J

c) The potential energy at from point Pis 0.19 J

d) The potential energy at point Q is 0.038 J

e) The potential energy at top of the loop is 0.075 J

f) If some initial speed is given to the block i.e.,vi0 then potential energy or work does not change, hence, there will be no change in the result of a) and e).

Step by step solution

01

Given

i) Mass of block m=0.032 kg

ii) Loop radius R=12 cm=12×102 m

iii) Gravitational acceleration g=9.8m/s2

iv) Height of block from the bottom h=5R=60×102m

02

To understand the concept

The force of gravity is constant, so the work done can be found by using the formula for the work done in terms of gravitational force and displacement.

Formula:

Gravitational potential energy is given bytheformula,U=mgh

Gravitational force is mgF=mg

03

(a) Calculate how much work does the gravitational force do on the block as the block travels from point P to point Q

Work depends on displacement.

Displacement from pointPto QPointis

d=hR

d=5RR

d=4R

Wg=mgd

Wg=mg4R

Wg=0.032×9.8×4×12×102

Wg=0.15 J

04

(b) Calculate how much work does the gravitational force do on the block as the block travels from point P to the top of the loop

Work depends on displacement. Top loop is from 2R distance fromthebottom.

Displacement from pointto top of the loop is

d=h2R

d=5R2R

d=3R

Wg=mgd

Wg=mg4R

Wg=0.032×9.8×3×12×102

Wg=0.11 J

05

 Step 5: (c) Calculate potential energy when the block is at point Pif the gravitational potential energy of the block–Earth system is taken to be zero at the bottom of the loop  

The potential energy at pointis

U=mgh

U=5mgR

U=0.032×9.80×5×12×102

U=0.19 J

06

(d) Calculate potential energy when the block is at point Pif the gravitational potential energy of the block–Earth system is taken to be zero at point Q 

At point Q total height isR

The potential energy at pointQis

U=mgR

U=0.032×9.80×12×102

U=0.038 J

07

(e) Calculate potential energy when the block is at point Pif the gravitational potential energy of the block–Earth system is taken to be zero at the top of the loop

At top of the loop total height is

The potential energy at top of the loop is

U=2mgR

U=0.032×9.80×12×102

U=0.075 J

08

(f) Figure out do the answers to (a) through (e) increase, decrease, or remain the same if of merely being released, the block is given some initial speed downward along the track 

The potential energy or work does not depend on the initial speed so, if initial speed Vi0then the result a) and e) is unchanged.

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