Q#9
One person says that the potential energy of a particular book kept in an almirah is 20 J and the other says it is 30 J. Is one of them necessarily wrong?
Answer:
No. One is free to choose the zero potential energy configuration as the origin is chosen arbitrarily. Hence the potential energy depends on this chosen configuration.
Q#10
A book is lifted from the floor and is kept in an almirah. One person says that the potential energy of the book is increased by 20 J and the other says it is increased by 30 J. Is one of them necessarily wrong?
Answer:
Yes. Because in this case difference in potential energy depends on the height of the almirah which is fixed. So only one person is right and the other is necessarily wrong.
Q#11
In one of the exercises to strengthen the wrist and fingers, a person squeezes and releases a soft rubber ball. Is the work done on the ball positive, negative or zero during compression? During expansion?
Answer:
Since the movement is along the force during compression the work done is positive. During expansion movement is opposite so the work done on the ball is negative.
Q#12
In a tug of war, the team that exerts a larger tangential force on the ground wins. Consider the period in which a team is dragging the opposite team by applying a larger tangential force on the ground. List which of the following works are positive, which are negative and which are zero?
(a) work by the winning team on the losing team
(b) work by the losing team on the winning team
(c) work by the ground on the winning team
(d) work by the ground on the losing team
(e) total external work done on the two teams.
(b) work by the losing team on the winning team
(c) work by the ground on the winning team
(d) work by the ground on the losing team
(e) total external work done on the two teams.
Answer:
(a) Positive
(b) Negative
(c) Zero (Assuming that work by ground means work by frictional force by ground, Also assuming the winning team does not step backwards)
(b) Negative
(c) Zero (Assuming that work by ground means work by frictional force by ground, Also assuming the winning team does not step backwards)
(d) Negative (Assuming that work by ground means work by frictional force by ground, Losing team is dragged)
(e) Zero (Taking rope, both teams and ground as a system, no external force is on the system, so no external work done)
Q#13
When an apple falls from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground?
Answer:
Since mechanical energy is conserved when no external force is applied, in this case, the gravitational potential energy is converted to kinetic energy just as it reaches the ground. After it strikes the ground, a part of this energy is used to deform the apple and other is converted to sound and heat energy.
Q#14
When you push your bicycle up on an incline the potential energy of the bicycle and yourself increases. Where does this energy come from?
Answer:
Chemical energy (converted from food) stored in our muscle is converted to mechanical energy that increases the P.E. of myself with the bicycle. The energy in the food is received from sunlight.
Q#15
The magnetic force on a charged particle is always perpendicular to its velocity. Can the magnetic force change the velocity of the particle? Speed of the particle?
Answer:
Since the magnetic force on the particle is perpendicular to its velocity, it can not change the speed of the particle but it can change its direction. Due to change in the direction, the velocity changes.
Q#16
A ball is given a speed v0 on a rough horizontal surface. The ball travels through a distance l on the surface and stops. (a) what are the initial and final kinetic energies of the ball? (b) what is the work done by the kinetic friction?
Answer:
(a) Initial Kinetic energy = ½mv0²
Final Kinetic Energy = 0
(b) Forces on the ball are weight, normal force and kinetic friction. Since weight and normal force are equal and opposite, the resultant force on the ball is equal to kinetic friction. Work done by the resultant force is equal to change in kinetic energy, so work done by kinetic friction
W = 0 - ½mv0² = -½mv0²
Q#17
Consider the situation of the previous question from a frame moving with a speed v0 parallel to initial velocity of the block. (a) What are the initial and final kinetic energies? (b) what is the work done by the kinetic friction?
Consider the situation of the previous question from a frame moving with a speed v0 parallel to initial velocity of the block. (a) What are the initial and final kinetic energies? (b) what is the work done by the kinetic friction?
Answer:
(a) In this moving frame initial velocity becomes zero, so initial K.E. =0, Final velocity = -v0, So final K.E. = ½mv0²
(b) work done by kinetic friction = ½mv0²
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