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

Uis defined as,

U=mghHere,

mis the mass of the body,gis the free fall acceleration andhis the fall of height.

Solution:-To find the potential energy of the object at the point

A, substitute 234 g for mass of the objectm, 9.81 m/s^{2}for free fall accelerationgand 1.05 m for heighthin the equationU=mgh,

U=mgh= (234 g) (9.81 m/s

^{2}) (1.05 m)= (234 g×10

^{-3}kg/1 g) (9.81 m/s^{2}) (1.05 m)= 2.41 kg. m

^{2}/s^{2}= (2.41 kg. m

^{2}/s^{2}) (1 J/1 kg. m^{2}/s^{2})= 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 pointAto 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

nsignifies 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 e

^{2}/r^{2},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 r

_{1}about the proton and jumps suddenly into a circular orbit of smaller radius r_{2 }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

Kof a body is defined as,

K= ½mv^{2}Here

mis the mass of the body andvis the velocity of the body.In a circular motion, the centripetal force

F_{c}acting on the particle is defined as,

F_{c}=mv^{2}/rHere

mis the mass particle,vis the velocity of the particle andris 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 (

F_{c}=mv^{2}/r) must be equal to the force of attraction (F=ke^{2}/r^{2}) between the positively charged proton and the negatively charged electron in the hydrogen atom.So,

F_{c}=F

mv^{2}/r=ke^{2}/r^{2}Or,

mv^{2}=ke^{2}/rThe potential energy difference

Uis defined as,

U= -∫FdrHere

Fis the force.The total energy Δ

Ewill be,Δ

E= ΔK+ ΔU

Solution:-(a) The kinetic energy

Kof the electron will be,

K= ½mv^{2}= ½ (

ke^{2}/r) (Sincemv^{2}=ke^{2}/r)Since the electron is initially moving in a circle of radius

r_{1}about the proton and jumps suddenly into a circular orbit of smaller radiusr_{2}, therefore the change in kinetic energy ΔKwill be,Δ

K= ½ke^{2}(1/r_{2}– 1/r_{1})From the above observation we conclude that, the change in kinetic energy of the electron Δ

Kwill be ½ke^{2}(1/r_{2}– 1/r_{1}).(b) To obtain the change in potential energy ΔU, substituteke^{2}/r^{2}for F in the equationU= -∫Fdrand integrating from the limitr_{1}tor_{2},Thus the change in potential energy Δ

Uof the atom will be -ke^{2}(1/r_{2}– 1/r_{1}).c) To obtain the total energy Δ

Eof the atom changed in this process, substitute ½ke^{2}(1/r_{2}– 1/r_{1}) for ΔKand -ke^{2}(1/r_{2}– 1/r_{1}) for ΔUin the equation ΔE= ΔK+ ΔU,Δ

E= ΔK+ ΔU= (½

ke^{2}(1/r_{2}– 1/r_{1})) + (-ke^{2}(1/r_{2}– 1/r_{1}))= -(½

ke^{2}(1/r_{2}– 1/r_{1}))From the above observation we conclude that, the total energy Δ

Eof the atom changed in this process would be -(½ke^{2}(1/r_{2}– 1/r_{1})).

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

Fhaving potential energyU(x) will be,

F= -dU(x)/dxKinetic energy

Kof a particle is defined as,

K= ½mv^{2}Here, mass of the particle is

mand speed of the particle isv.So the velocity of the particle will be,

v= √2K/mTotal energy

Eis equal to the sum of kinetic energyKand potential energyU.

E=K+USo 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

xaxis through a region in which its potential energyU(x) varies.Using figure,

Δ

U= -17 J – (-3 J)= -14 J

Δ

x= 4 m-1 m= 3 m

To obtain the force acting on the particle

Fatx= 2.0 m, substitute -14 J for ΔUand 3 m for Δxin the equationF= -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

Fatx= 2.0 m would be 4.7 N.(b) To find the kinetic energy of the particle at atx= 2.0 m, substitute 2.0 kg for massmand -2.0 m/s forvin the equationK= ½mv^{2},

K= ½mv^{2}=1/2 (2.0 kg) (-2.0 m/s)

^{2}= 4 kg.m

^{2}/s^{2}= (4 kg.m

^{2}/s^{2}) (1 J/1 kg.m^{2}/s^{2})= 4 J

From the figure, the potential energy

Uis,

U= -7 JSo the total energy

Eis equal to the sum of kinetic energyKand potential energyU.

E=K+U= 4 J+ (-7 J)

= -3 J

This signifies that, the particle is constrained to move between

x= 1 m andx= 14 m.(c) For

x= 7 m, the total energy isE= -3 J and potential energyU= -17 J.To obtain the kinetic energy

K, substitute -3 J forEand -17 J forUin the equationK=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 forKand 2.0 kg formin the equationv= √2K/m,

v= √2K/m= √2(14 J) /(2.0 kg) ((1 kg.m

^{2}/s^{2})/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 (

K_{r}) of a flywheel is defined as,

K_{r}= ½Iw^{2}…… (1)Here

Iis the rotational inertia of the flywheel andwis the angular velocity of the flywheel.Momentum of inertia (

I) of the flywheel is defined as,

I= ½mr^{2}…… (2)To obtain the rotational kinetic energy (

K_{r}) in terms of mass (m) and radius (r), substitutemr^{2}forIin the equationK_{r}= ½Iw^{2},

K_{r}= ½Iw^{2}= ½ (1/2

mr^{2})(w^{2})= ¼ (

mr^{2})(w^{2}) …… (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 (

K_{r}) of the flywheel after charging, substitute 512 kg for massm, 97.6 cm for radius r and 624 rad/s for angular velocitywin the equationK_{r}= ¼ (mr^{2})(w^{2}),

K_{r}= ¼ (mr^{2})(w^{2})= ¼ (512 kg) ((97.6 cm)(10

^{-2}m/1 cm))^{2}(624 rad/s)^{2}= 4.75×10

^{7}kg. m^{2}/s^{2}= (4.75×10

^{7}kg. m^{2}/s^{2}) (1 J/ kg. m^{2}/s^{2})= 4.75×10

^{7}J …… (6)From equation (6) we observed that, the rotational kinetic energy (

K_{r}) of the flywheel after charging will be 4.75×10^{7}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 4.75×10^{7}J for workWand 8.13 kW for powerPin the equationt=W/P,

t=W/P= 4.75×10

^{7}J / 8.13 kW= (4.75×10

^{7}J / 8.13 kW) (1 kW/ 10^{3}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

W_{g}is defined as,

W_{g}=mghHere

mis the mass,gis the free fall acceleration andhis the height.Work done by the force of spring

W_{s}is defined as,

W_{s}= ½kx^{2}Here

kis the force constant andxis the extension.Kinetic energy

Kof a body is defined as,

K= ½mv^{2}Here

mis the mass andvis the speed of the object.So,

v=√2K/m

Solution:-

(a)Here the work done by the force of gravityW_{g}will be,

W_{g}=m(-g)h= -

mghTo obtain the work done by the force of gravity

W_{g}, substitute 263 g form, 9.81 m/s^{2}forgand -11.8 cm forhin the equationW_{g}= -mgh,

W_{g}= -mgh= -(263 g)(9.81 m/s

^{2})(-11.8 cm)= -(263 g(10

^{-3}kg/1 g))(9.81 m/s^{2})(-11.8 cm(10^{-2}m/1 cm))= (0.304 kg.m

^{2}/s^{2}) (1 J/1 kg.m^{2}/s^{2})= 0.304 J

Thus the work done by the force of gravity

W_{g}would be 0.304 J.

(b)As the spring is being compressed, the work done by the force of springW_{s}will be,

W_{s}= -½k(-x)^{2}To obtain the work done by the force of spring

W_{s}, substitute 2.52 N/cm fork, and -11.8 cm forxin the equationW_{s}= -½k(-x)^{2},

W_{s}= -½k(-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

W_{s}would be -1.75 J.

(c)The kinetic energyKjust before hitting the block would be,

K= 1.75 J-0.304 J= 1.45 J

To obtain the speed

vof the block just before it hit the spring, substitute 1.45 J for K and 263-g formin the equationv=√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.m

^{2}/s^{2}/1 J)/(0.263 kg)=3.32 m/s

Thus the speed

vof 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)

y^{2}-(0.263 kg)(9.81 m/s^{2})y½ (252 N/m)

y^{2}+(0.263 kg)(9.81 m/s^{2})y-5.78 J=0(252 N/m)

y^{2}+(5.16 kg.m/s^{2})y-11.56 J =0The 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/tSo the energy

Wwill be,

W= (P) (t)So the time (rate)

twill be equal to,

t=W/P

Solution:-

(a)First we have to convert kilowatt-hours to joules.1 kW.h = (1 kW) (10

^{3}W/1 kW) (1 h) (60 min/1 h) (60 s/1 min)= (3.6×10

^{6}W.s) (1 J/1 W.s)= 3.6×10

^{6}JThe car gets 30 mi/gal, and one gallon of gas produces 140 MJ of energy.

The gas required to produce 3.6×10

^{6}J will be,3.6×10

^{6}J (1 gal/140×10^{6}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 timetto 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 3.6×10^{6}J for energyWand 51 s for time t in the equationP=W/t,

P=W/t= (3.6×10

^{6}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 v

_{0}required for the particle to just reach the top of the bowl. Find v_{0}as a function ofθ_{0, }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 (

K_{i}+U_{i}) is equal to the final value (K_{f}+U_{f}). HereK_{i}is the initial kinetic energy,U_{i}is the initial potential energy,K_{f}is the finalkinetic energy and

U_{f}is the final potential energy.So,

K_{f}+U_{f}=K_{i}+U_{i}½

mv_{f}^{2}+mgy_{f}= ½mv_{i}^{2}+mgy_{i}Here

mis the mas of the body,v_{i}is the initial velocity,v_{f}is the final velocity,y_{i}is the initial height andy_{f}is the final height andgis the acceleration due to gravity.

Solution:-Initially the particle has the kinetic energy

K_{i}(=1/2mv_{0}^{2}) and potential energy isU_{i}(=mgy). Finally the particle has the kinetic energy at the top of the bowl is zero (K_{f}=0) and the potential energy is also zero (U_{f }= 0) at rest position, since we are taking horizontal axis as our reference point.So,

U_{i}= 0Thus,

K_{i}+U_{i}=K_{f}+U_{f}= 0+0

=0

1/2

mv_{0}^{2}+mgy=0So,

v_{0}= √-2gyHere

yis the distance beneath the rim and applying geometry it will be equal to –rcosθ_{0}. Negative sign signifies the downward direction.To find out initial speed

v_{0}of the particle, substitute –rcosθ_{0}foryin the equationv_{0}= √-2gy,

v_{0}= √-2gy= √-2

g(–rcosθ_{0})= √2

g(rcosθ_{0})From the above observation we conclude that, out initial speed

v_{0}of the particle would be √2g(rcosθ_{0}).

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.

Related Resources:-

- Click here for the Detailed Syllabus of IIT JEE Physics.
- Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.
- You can get the knowledge of Useful Books of Physics.

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