In the figure, a small block of mass m 0 048 kg can slide along the frictionless

Question

In the figure, a small block of mass m 0 048 kg can slide along the frictionless loop-the-loop, with loop radius R = 20 cm. The blockis released from rest at port P at height h 9R 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 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 , and (e) at the top of the loop? QT

Explanation / Answer

a)Work done by the gravitation force on the body = increase in its kinetic energy = decrease in its potential energy

so, decrease in its potential energy = mgHi - mgHf

= mg*(9R-R)

=8*mgR

=8*0.048*9.81*0.2

=0.753 J

so work done by Gravitaiton force = 0.753 J

b)using the similar procedure as above,

Work done = loss in potential energy

=mgHi - mgHf

=9mgR - 2mgR

=7mgR

=7*0.048*9.81*0.2

=0.659 J

c)Potential energy = mgH

=mg*9*R

=9mgR

=9*0.048*9.81*0.2

=0.847 J

d)PE = mgH

=mgR

=0.048*9.81*0.2

=0.094 J

e)PE = mgH

=2mgR

=2*0.048*9.81*0.2

=0.188 J

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