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NEET-XII-Physics

08: Work and Energy

with Solutions - page 2

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  • #12-b
    work by the losing team on the winning team
    Ans : Work by the losing team on the winning team is negative, as the displacement of the winning team is opposite to the force applied by losing team.
  • #12-c
    work by the ground on the winning team
    Ans : Work by the ground on the winning team is positive.
  • #12-d
    work by the ground on the losing team
    Ans : Work by the ground on the losing team is negative.
  • #12-e
    total external work on the two teams.
    Ans : Total external work on the two teams is positive.
    Page No 131:
  • Qstn #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?
    Ans : When an apple falls from a tree, its gravitational potential energy decreases as it reaches the ground. After it strikes the ground, its potential energy will remain unchanged.
    Page No 131:
  • Qstn #14
    When you push your bicycle up on an inclined plane, the potential energy of the bicycle and yourself increases. Where does this energy come from?
    Ans : When a person pushes his bicycle up on an inclined plane, the potential energies of the bicycle and the person increase because moving up on the inclined plane the kinetic energy decreases. and as mechanical energy is sum of kinetic energy and potential energy, and remains constant for a conservative system. Therefore, potential energy must increase in this case.
    Page No 131:
  • Qstn #15
    The magnetic force on a charged particle is always perpendicular to its velocity. Can the magnetic force change the velocity of the particles? Speed of the particle?
    Ans : The magnetic force on a charged particle is always perpendicular to its velocity. Therefore, the work done by the magnetic force on the charged particle is zero. Here, the kinetic energy and speed of the particle remain unaffected, while the velocity changes due to the change in direction of its motion.
    Page No 131:
  • Qstn #16
    A ball is given a speed v on a rough horizontal surface. The ball travels through a distance l on the surface and stops.
  • #16-a
    what are the initial and final kinetic energies of the ball?
    Ans : Initial kinetic energy of the ball, `` {K}_{i}=\frac{1}{2}m{v}^{2}``
    Here, m is the mass of the ball.
    The final kinetic of the ball is zero.
  • #16-b
    What is the work done by the kinetic friction?
    Ans : Work done by the kinetic friction is equal to the change in kinetic energy of the ball.
    ∴ Work done by the kinetic friction = `` {K}_{f}-{K}_{i}=0-\frac{1}{2}m{v}^{2}``
    = `` -\frac{1}{2}m{v}^{2}``
    Page No 131:
  • Qstn #17
    Consider the situation of the previous question from a frame moving with a speed v0 parallel to the initial velocity of the block.
    Ans : The relative velocity of the ball w.r.t. the moving frame is given by `` {v}_{r}=v-{v}_{0}``.
  • #17-a
    What are the initial and final kinetic energies?
    Ans : Initial kinetic energy of the ball = `` \frac{1}{2}m{{v}_{r}}^{2}=\frac{1}{2}m(v-{v}_{0}{)}^{2}``
    Also, final kinetic energy of the ball = `` \frac{1}{2}m(0-{v}_{0}{)}^{2}=\frac{1}{2}m{{v}_{0}}^{2}``
  • #17-b
    What is the work done by the kinetic friction?
    Ans : Work done by the kinetic friction = final kinetic energy `` -`` initial kinetic energy
    = `` \frac{1}{2}m({v}_{0}{)}^{2}-\frac{1}{2}m(v-{v}_{0}{)}^{2}``
    = `` -\frac{1}{2}m{v}^{2}+mv{v}_{0}``
    Page No 131:
  • #
    Section : ii
  • Qstn #1
    A heavy stone is thrown from a cliff of height h with a speed v. The stoen will hit the ground with maximum speed if it is thrown
    (a) vertically downward
    (b) vertically upward
    (c) horizontally
    (d) the speed does not depend on the initial direction.
    digAnsr:   d
    Ans : (d) the speed does not depend on the initial direction
    As the stone falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.
    From the conservation of energy, we have:
    Initial energy of the stone = final energy of the stone
    `` \,\mathrm{\,i\,}.\,\mathrm{\,e\,}.,(K.E.{)}_{i}+(P.E.{)}_{i}=(K.E.{)}_{f}+(P.E.{)}_{f}``
    `` \Rightarrow \frac{1}{2}m{v}^{2}+mgh=\frac{1}{2}m({v}_{max}{)}^{2}``
    `` \Rightarrow {v}_{max}=\sqrt{{v}^{2}+2gh}``
    `` ``
    From the above expression, we can say that the maximum speed with which stone hits the ground does not depend on the initial direction.
    Page No 131:
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