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• Complete Physics Course - Class 11
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Solved Examples on Work, Power and Energy:-

Problem 1:-

A small object of mass m = 234 g slides along a track with elevated ends and a central flat part, as shown in below figure. The flat part has a length L = 2.16 m. The curved portions the tracks are frictionless; but in traversing the flat part, the object loses 688 mJ of mechanical energy, due to friction. The object is released at point A, which is a height h = 1.05 m above the flat part of the track. Where does the object finally come to rest?

Concept:-

The potential energy U is defined as,

U = mgh

Here, m is the mass of the body, g is the free fall acceleration and h is the fall of height.

Solution:-

To find the potential energy of the object at the point A, substitute 234 g for mass of the object m, 9.81 m/s2 for free fall acceleration g and 1.05 m for height h in the equation U = mgh,

U = mgh

= (234 g) (9.81 m/s2) (1.05 m)

= (234 g×10-3 kg/1 g) (9.81 m/s2) (1.05 m)

= 2.41 kg. m2/s2

= (2.41 kg. m2/s2) (1 J/1 kg. m2/s2)

= 2.41 J

The curved portions of the track are frictionless; but in traversing the flat part, the object losses 688 mJ of mechanical energy, due to friction.

The number of times (n)  that the particle will move back and forth across the flat portion is the ratio of the potential energy (2.41 J) of the object at the point A to the mechanical energy (688 mJ)that the object losses due to friction.

So, n = 2.41 J/688 mJ

= (2.41 J/688 mJ) (1 mJ/10-3 J)

= 3.50

The number of times n signifies that, the particle will come to a rest at the center of the flat part while attempting one last right to left journey.

Problem 2:-

The magnitude of the force of attraction between the positively charged proton and the negatively charged electron in the hydrogen atom is given by,

F = K e2/r2,

where e is the electric charge of the electron, k is a constant, and r is the separation between electron and proton. Assume that the proton is fixed. Imagine that the electron is initially moving in a circle of radius r1 about the proton and jumps suddenly into a circular orbit of smaller radius r2 as shown in the below figure. (a) Calculate the change in kinetic energy of the electron, using Newton’s second law. (b) Using the relation between force and potential energy, calculate the change in potential energy of the atom. (c) By how much has the total energy of the atom changed in this process? (This energy is often given off in the form of radiation.)

Concept:-

Kinetic energy K of a body is defined as,

K = ½ mv2

Here m is the mass of the body and v is the velocity of the body.

In a circular motion, the centripetal force Fc acting on the particle is defined as,

Fc = mv2/r

Here m is the mass particle, v is the velocity of the particle and r is the radius of the circle.

Apply Newton’s second law to the motion of electron in a circular orbit of the hydrogen atom where the proton is at the center of circle.

So the centripetal force (Fc = mv2/r) must be equal to the force of attraction (F= ke2/r2) between the positively charged proton and the negatively charged electron in the hydrogen atom.

So, Fc = F

mv2/r = ke2/r2

Or, mv2 = ke2/r

The potential energy difference U is defined as,

U = -∫ Fdr

Here F is the force.

The total energy ΔE will be,

ΔE = ΔK + ΔU

Solution:-

(a) The kinetic energy K of the electron will be,

K = ½ mv2

= ½ (ke2/r)       (Since mv2 = ke2/r)

Since the electron is initially moving in a circle of radius r1 about the proton and jumps suddenly into a circular orbit of smaller radius r2, therefore the change in kinetic energy ΔK will be,

ΔK = ½ ke2 (1/r2 – 1/r1)

From the above observation we conclude that, the change in kinetic energy of the electron ΔK will be ½ ke2 (1/r2 – 1/r1).

(b) To obtain the change in potential energy ΔU, substitute ke2/r2 for F in the equation U = -∫ Fdr and integrating from the limit r1 to r2,$U-_int_r_1r_2Fdr.jpg$

Thus the change in potential energy ΔU of the atom will be -ke2 (1/r2 – 1/r1).

c) To obtain the total energy ΔE of the atom changed in this process, substitute ½ ke2 (1/r2 – 1/r1) for ΔK and -ke2 (1/r2 – 1/r1) for ΔU in the equation ΔE = ΔK + ΔU,

ΔE = ΔK + ΔU

= (½ ke2 (1/r2 – 1/r1)) + (-ke2 (1/r2 – 1/r1))

= -(½ ke2 (1/r2 – 1/r1))

From the above observation we conclude that, the total energy ΔE of the atom changed in this process would be -(½ ke2 (1/r2 – 1/r1)).

Problem 3:-

A particle of mass 2.0 kg moves along the x axis through a region in which its potential energy U(x) varies as shown in below figure. When the particle is at x = 2.0 m, its velocity is -2.0 m/s. (a) Calculate the force acting on the particle at this position. (b) Between what limits does the motion takes place? (c) How fast is it moving when it is at x = 7.0 m?

Concept:-

Force acting on the particle F having potential energy U(x) will be,

F = - dU(x)/dx

Kinetic energy K of a particle is defined as,

K = ½ mv2

Here, mass of the particle is m and speed of the particle is v.

So the velocity of the particle will be,

v = √2K/m

Total energy E is equal to the sum of kinetic energy K and potential energy U.

E = K+U

So the kinetic energy K will be,

K = E-U

Solution:-

(a) The above figure shows a particle of mass 2.0 kg moves along the x axis through a region in which its potential energy U(x) varies.

Using figure,

ΔU = -17 J – (-3 J)

= -14 J

Δx = 4 m-1 m

= 3 m

To obtain the force acting on the particle F at x = 2.0 m, substitute -14 J for ΔU and 3 m for Δx in the equation F = - dU(x)/dx,

F = - dU(x)/dx

= - (-14 J/3 m)

= (4.7 J/m) (1 N/(1 J/m))

= 4.7 N

From the above observation we conclude that, force acting on the particle F at x = 2.0 m would be 4.7 N.

(b) To find the kinetic energy of the particle at at x = 2.0 m, substitute 2.0 kg for mass m and -2.0 m/s for v in the equation K = ½ mv2,

K = ½ mv2

=1/2 (2.0 kg) (-2.0 m/s)2

= 4 kg.m2/s2

= (4 kg.m2/s2) (1 J/1 kg.m2/s2)

= 4 J

From the figure, the potential energy U is,

U = -7 J

So the total energy E is equal to the sum of kinetic energy K and potential energy U.

E = K+U

= 4 J+ (-7 J)

= -3 J

This signifies that, the particle is constrained to move between x = 1 m and x = 14 m.

(c) For x= 7 m, the total energy is E = -3 J and potential energy U = -17 J.

To obtain the kinetic energy K, substitute -3 J for E and -17 J for U in the equation K = E-U, we get,

K = E-U

= (-3 J) – (-17 J)

= 14 J

To obtain the speed of the particle when it is at x = 7.0 m, substitute 14 J for K and 2.0 kg for m in the equation v = √2K/m,

v = √2K/m

= √2(14 J) /(2.0 kg)  ((1 kg.m2/s2)/1 J)

= 3.7 m/s

From the above observation we conclude that, the speed of the particle when it is at x = 7.0 m would be 3.7 m/s.

Problem 4:-

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are changed by using an electric motor to get the flywheel up to its top speed of 624 rad/s. One such fly wheel is a solid, homogeneous cylinder with a mass of 512 kg and a radius of 97.6 cm. (a) What is the kinetic energy of the flywheel after charging? (b) If the truck operates with an average power requirement of 8.13 kW, for how many minutes can it operate between charging’s?

Concept:-

Rotational kinetic energy (Kr) of a flywheel is defined as,

Kr = ½ Iw2      …… (1)

Here I is the rotational inertia of the flywheel and w is the angular velocity of the flywheel.

Momentum of inertia (I) of the flywheel is defined as,

I = ½ mr2      …… (2)

To obtain the rotational kinetic energy (Kr) in terms of mass (m) and radius (r), substitute mr2 for I  in the equation Kr = ½ Iw2,

Kr = ½ Iw2

= ½ (1/2 mr2)(w2)

= ¼ (mr2)(w2) …… (3)

The power (P) due to the force acting on a body is equal to the work done on the body (W) divided by time (t).

P = W/t          …… (4)

From equation (4),  the time (t) will be,

t = W/P        …… (5)

Solution:-

(a) To obtain the rotational kinetic energy (Kr) of the flywheel after charging, substitute 512 kg for mass m, 97.6 cm for radius r and 624 rad/s for angular velocity w in the equation Kr = ¼ (mr2)(w2),

Kr = ¼  (mr2)(w2)

= ¼  (512 kg) ((97.6 cm)(10-2 m/1 cm)) 2(624 rad/s) 2

= 4.75×107 kg. m2/s2

= (4.75×107 kg. m2/s2) (1 J/ kg. m2/s2)

= 4.75×107 J      …… (6)

From equation (6) we observed that, the rotational kinetic energy (Kr) of the flywheel after charging will be 4.75×107 J.

(b) We have to find out the time in minutes which can operate between charging’s, when the truck operates with an average power requirement of 8.13 kW.

To obtain time (t), substitute 4.75×107 J for work W and 8.13 kW for power P in the equation t = W/P,

t = W/P

= 4.75×107 J / 8.13 kW

= (4.75×107 J / 8.13 kW) (1 kW/ 103 W)

= 5842.6 J/W

= (5842.6 J/W) (1 s/(1 J/W))

= 5842.6 s

= (5842.6 s) ((1/60 minutes)/ 1 s)

= 97.4 minutes        …… (7)

From equation (7) we observed that, the truck can operate 97.4 minutes between charging’s.

Problem 5:-

A 263-g block is dropped onto a vertical spring with force constant k = 2.52N/cm as shown in the below figure. The block sticks to the spring, and the spring compress 11.8 cm before coming momentarily to rest. while the spring is being compressed, how much work is done (a) by the force of gravity and (b) by the spring? (c)  What was the speed of the block just before it hit the spring? (d)  If this initial speed of the block is doubled, what is the maximum compression of the spring? Ignore friction.

Concept:

Work done by the force of gravity Wg is defined as,

Wg = mgh

Here m is the mass, g is the free fall acceleration and h is the height.

Work done by the force of spring Ws is defined as,

Ws = ½ kx2

Here k is the force constant and x is the extension.

Kinetic energy K of a body is defined as,

K = ½ mv2

Here m is the mass and v is the speed of the object.

So, v=√2K/m

Solution:-

(a) Here the work done by the force of gravity Wg will be,

Wg = m(-g)h

= -mgh

To obtain the work done by the force of gravity Wg, substitute 263 g for m, 9.81 m/s2 for g and -11.8 cm for h in the equation Wg= -mgh,

Wg= -mgh

= -(263 g)(9.81 m/s2)(-11.8 cm)

= -(263 g(10-3 kg/1 g))(9.81 m/s2)(-11.8 cm(10-2 m/1 cm))

= (0.304 kg.m2/s2) (1 J/1 kg.m2/s2)

= 0.304 J

Thus the work done by the force of gravity Wg would be 0.304 J.

(b) As the spring is being compressed, the work done by the force of spring Ws will be,

Ws = -½ k (-x)2

To obtain the work done by the force of spring Ws, substitute 2.52 N/cm for k, and -11.8 cm for x in the equation Ws = -½ k (-x)2,

Ws = -½ k (-x)2

= - ½ (2.52 N/cm)(-11.8 cm)2

= - ½ (2.52 N/cm) (100 cm/1 m)(-11.8 cm (10-2 m/1 cm))2

= (-1.75 N.m) (1 J/1 N.m)

= -1.75 J

Thus the work done by the force of spring Ws would be -1.75 J.

(c) The kinetic energy K just before hitting the block would be,

K = 1.75 J-0.304 J

= 1.45 J

To obtain the speed v of the block just before it hit the  spring, substitute 1.45 J for K and 263-g for m in the equation v=√2K/m,

v=√2K/m

= √2(1.45 J)/(263 g)

=√2(1.45 J)/(263 g (10-3 kg/1 g))

=√2(1.45 J)/(0.263 kg)

=√2(1.45 J) (1 kg.m2/s2/1 J)/(0.263 kg)

=3.32 m/s

Thus the speed v of the block would be 3.32 m/s.

(d) As the initial speed of the block is doubled, the initial kinetic energy term will be quadruples to 5.78 J.

The compression will then be given by,

-5.78 J= - ½ (252 N/m)y2-(0.263 kg)(9.81 m/s2)y

½ (252 N/m)y2+(0.263 kg)(9.81 m/s2)y -5.78 J=0

(252 N/m)y2+(5.16 kg.m/s2)y-11.56 J =0

The above equation is a quadratic equation. The solution of the above equation will be,

y = [[-5.16±√(5.16)2-4(252)(-11.56)]/2(252)] m

= 0.225 m

From the above observation we conclude that, the maximum compression of the spring would be 0.225 m.

Problem 6:-

Suppose that your car averages 30 mi/gal of gasoline. (a) How far could you travel on 1 kW.h of energy consumed? (b) If you are driving at 55 mi/h, at what rate are you expending energy? The heat of combustion of gasoline is 140 MJ/gal.

Concept:-

Power (P) is equal to the energy (W) divided by time (t).

P = W/t

So the energy W will be,

W = (P) (t)

So the time (rate) t will be equal to,

t = W/P

Solution:-

(a) First we have to convert kilowatt-hours to joules.

1 kW.h = (1 kW) (103 W/1 kW) (1 h) (60 min/1 h) (60 s/1 min)

= (3.6×106 W.s) (1 J/1 W.s)

= 3.6×106 J

The car gets 30 mi/gal, and one gallon of gas produces 140 MJ of energy.

The gas required to produce 3.6×106 J will be,

3.6×106 J (1 gal/140×106 J) = 0.026 gal.

Thus the distance traveled on this much gasoline will be,

(0.026 gal) (30 mi/1 gal) = 0.78 mi

From the above observation we conclude that, you could travel 0.78 mi on 1 kW.h of energy consumed.

(b) As you are driving at 55 mi/h, therefore the time t to travel 0.78 mi will be,

t = (0.78 mi) (1 h/55 mi)

= 0.014 h

= 51 s

To obtain the rate of energy expenditure (P), substitute  3.6×106 J for energy W and 51 s for time t in the equation P = W/t,

P = W/t

= (3.6×106 J)/(51 s)

= (71, 000 J/s) (1 W/(1J/s))

= 71,000 W

From the above observation we conclude that, the rate of energy expenditure (P) would be 71,000 W.

Problem 7:-

A particle is projected horizontally along the interior of a frictionless hemispherical bowl of radius r, which is kept at rest in below figure. We wish to find the initial speed v0 required for the particle to just reach the top of the bowl. Find v0 as a function of θ0, the initial angular position of the particle.

Concept:-

In an isolated system in which only conservative forces act, the total mechanical energy remains constant. That is, the initial value of the total mechanical energy (Ki +Ui)  is equal to the final value (Kf +Uf). Here Ki is the initial kinetic energy, Ui is the initial potential energy, Kf is the final

kinetic energy and Uf is the final potential energy.

So, Kf +Uf = Ki +Ui

½ mvf2 + mgyf = ½ mvi2 + mgyi

Here m is the mas of the body, vi is the initial velocity, vf is the final velocity, yi is the initial height and yf is the final height and g is the acceleration due to gravity.

Solution:-

Initially the particle has the kinetic energy Ki (=1/2 mv02) and potential energy is Ui (=  mgy). Finally the particle has the kinetic energy at the top of the bowl is zero (Kf=0) and the potential energy is also zero (Uf = 0) at rest position, since we are taking horizontal axis as our reference point.

So, Ui = 0

Thus, Ki +Ui  = Kf +Uf

= 0+0

=0

1/2 mv02+ mgy =0

So, v0 = √-2gy

Here y is the distance beneath the rim and applying geometry it will be equal to –r cos θ0. Negative sign signifies the downward direction.

To find out initial speed v0 of the particle, substitute –r cos θ0 for y in the equation v0 = √-2gy,

v0 = √-2gy

= √-2g(–r cos θ0)

= √2g(r cos θ0)

From the above observation we conclude that, out initial speed v0 of the particle would be √2g(r cos θ0).

Problem 8:-

A 1 kg block collides with a horizontal light spring of force constant 2 N/m. The block compresses the spring m from rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.25. What was the speed of the block at the instant of collision.

Solution:-

When the block compresses the spring let x be the amount of compression.

i.e. x = 4m

Let v be the velocity of the lock when it collides with the spring.

Loss in K.E. of the block = gain in elastic potential energy + work done against friction

(1/2)mv2 - 0 = (1/2)kx2 + µmgx,

(1/2)mv2 = (1/2) × 2 × 42 + 0.25 × 1 × 9.8 × 4,

v2 = 2 × 42 + 0.25 × 1 × 9.8 × 4 × 2 = 51.6,

v = √51.6 = 7.2m/sec

From the above observation, we conclude that, the speed of the block at the instant of collision would be 7.2 m/sec.

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