Work, Energy and Power
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6.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).
6.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J).
6.12 Electron mass, me = 9.11*10-31 kg
Proton mass, madhya pradesh = 1.67*10–27 kg
Electron's kinetic energy = 10 keV = 10 1.60 *10–19 J = 1.60 *10–15 J
Proton's kinetic energy = 100 keV = 100 1.60 *10–19 J = 1.60 *10–14 J
The electron kinetic energy is given by E
Upvote 
6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
F = − +3 N
where , , are unit vectors along the x-, y- and z-axis of the system respectively.
What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by
F = − +3 N
where , , are unit vectors along the x-, y- and z-axis of the system respectively.
What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
6.11: Force exerted on the body, F =? +3 N
Displacement, s = 4 km
Work done, W = F.s
= (? +3
= 0+0-3
= 12 J
Upvote 6.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) t 1/2
(ii) t
(iii) t 3/2
(iv) t 2
6.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) t 1/2
(ii) t
(iii) t 3/2
(iv) t 2
6.10 Power is given by the relation
P = Fv = mav = mv = constant ( say, k)
vdv =
v =
For displacement x of the body, we have:
v = =
dx = k' dt where k' = = constant
On integrating both sides, we get
x =
Therefore x
Upvote 6.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) t 1/2 (ii) t (iii) t 3/2 (iv) t 2
6.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) t 1/2 (ii) t (iii) t 3/2 (iv) t 2
6.9 Let us assume
Body mass = m
Acceleration = a
According to Newton's 2nd law F = MA (constant)
We know a = dv/dt = constant. Hence dv = dt constant
On integrating, v = t + constant
The relation of power is given by P = F
We have
Hence,
=
=
Therefore P
Upvote 6.8 Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic?
(Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
6.8 Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic?
(Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
6.8 (a) In elastic collision, the initial and final kinetic energy is equal. When the two balls collide, there is no conservation of kinetic energy; it gets converted into potential energy.
(b) The total linear momentum is conserved in an elastic collision.
(c) In case of inelastic cond
Upvote 6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system
6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system
6.7 (a) False. The total momentum and energy is conserved, not the individual.
(b) False. External forces can change the energy of a body.
(c) False. Only work done by conservative force over a closed loop is zero.
(d) True. In an inelastic collision, the final velocity reduc
Upvote 6.6 Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
6.6 Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
6.6 (a) When a conservative force does positive work on a body, the body gets displaced in the direction of the force, it moves towards the centre of the force, thus resulting in decrease of potential energy.
(b) When the work done by a body against friction, it reduces its velocity. Hence ki
Upvote 6.5 Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat
energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
6.5 Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat
energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
6.5 (a) As per the law of conservation of energy,
Total energy = potential energy + kinetic energy
= mgh + mv2
When the casing burns, mass reduces, resulting in drop of energy. Hence the energy for burning of casing is drawn from the rocket.
(b) The force due to gravity is a conserva
Upvote 6.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
6.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
6.4 Given, particle energy, E = 1 J,
Force constant, k = 0.5 N/m
Kinetic energy, KE = mv2
From the equation, total energy, E = KE +PE, we get
1 = (1/2)mv2 + (1/2) kx2
when it turns back, v becomes 0
1 = (1/2) , x = ± 2
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