Q:

Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

A:

Yes. When an object is moving with constant velocity, the net force acting on it is zero. However, there is a displacement along the motion of the object. Therefore, there can be a displacement of an object without any force acting upon it.

Q:

Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

A:

On pushing a huge rock, no muscular energy is transferred to the stationary rock. Here, muscular energy is transferred into heat energy which makes the body feel warm. Hence, there is no loss of energy.

Q:

What are the various energy transformations that occur when you are riding a bicycle?

A:

In the case of bicycle riding, the muscular energy of the rider changes into heat energy and kinetic energy of the bicycle. Heat energy gives heat to the rider’s body while kinetic energy provides a velocity to the bicycle. In this whole process, the total energy remains conserved.

Q:

Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

A:

The statement of Soni is correct. Acceleration in an object may be zero even when several forces are acting on it. This condition occurs when all the forces cancel out each other. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero.

Q:

A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

A:

When the person holds a bundle of hay over his head, then bundle of hay is not displaced. However, force of gravity is still working on the bundle but the person is not applying any force on it. Thus, work done by the person on the bundle is zero.

Q:

A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

A:

\begin{array}{l}\text{Here},\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{unit}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{kWh}\\ \text{Also},\text{\hspace{0.17em}}\text{1}\text{\hspace{0.17em}}\text{kWh}=3.6\times 1{0}^{6}\text{\hspace{0.17em}}\text{J}\\ \therefore \text{\hspace{0.17em}}\text{25}0\text{\hspace{0.17em}}\text{units}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{energy}\text{\hspace{0.17em}}=250\times 3.6\times 1{0}^{6}\text{J}=9\times 1{0}^{8}\text{J}\end{array}

Q:

The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

A:

No. When the body falls freely from a height, then its potential energy converts into kinetic energy with time. The decrease in the potential energy is equal to the increase in the kinetic energy of the body. Thus, total mechanical energy of the body remains conserved.

Q:

A battery lights a bulb. Describe the energy changes involved in the process.

A:

The chemical energy of battery is changed into electrical energy when a bulb is connected between the terminals of a battery. Later on, this electrical energy changes into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:

\text{Chemical}\text{\hspace{0.17em}}\text{energy}\text{\hspace{0.17em}}\to \text{\hspace{0.17em}}\text{Electrical}\text{\hspace{0.17em}}\text{Energy}\text{\hspace{0.17em}}\to \text{\hspace{0.17em}}\text{Light}\text{\hspace{0.17em}}\text{Energy}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\text{Heat}\text{\hspace{0.17em}}\text{Energy}

Q:

An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

A:

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement can be calculated by the difference in the initial and final heights of the object. Here, vertical displacement is zero. Hence, work done by gravity W = mgh

Here, h = 0

Therefore, W = mg × 0 = 0

Q:

A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

A:

In free fall, potential energy of an object decreases and kinetic energy increases. When the object comes in contact with ground, potential energy converts into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. The object can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.

Q:

Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.

A:

\begin{array}{l}\text{Given,Power of each device,P=500W=0}\text{.50kW}\\ \text{Time,t=10h}\\ \text{Energyconsumed=Power\xd7Time}\\ \text{=0}\text{.50kW\xd710 h=5 kWh}\\ \text{Hence,theenergyconsumedbyfourequalrating}\\ \text{devicesin10h=4\xd75 kWh=20units}\text{.}\end{array}

Q:

In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

A:

1. In first case, the direction of force acting on the block is perpendicular to the displacement. Thus, work done by force on the block will be zero.

2. In the second case, the direction of force acting on the block is in the direction of displacement. Hence, work done by force on the block will be positive.

3. In the third case, the direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block will be negative.

Q:

What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

A:

There are two conditions when the work is said to be done.

(i) A force acts on the body.

(ii) There is a displacement of the body due applied force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero.

When a satellite revolves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Thus, the work done on the satellite by the Earth is zero.

Q:

A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

A:

Work done by gravity depends only on the vertical displacement of the body. The expression for work done by gravity is given by,

W = mgh

Here, vertical displacement (h) is zero.

Hence, W = mg × 0 = 0

Thus, work done by gravity on the body is zero.

Q:

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

A:

\begin{array}{l}\text{Given,Mass,m=1500kg}\\ {\text{Velocity,v=60kmh}}^{\text{-1}}\text{=60\xd7}\frac{5}{18}{\text{ms}}^{\text{-1}}\\ \text{Theworkdonetostopthecarwouldbeequaltothekineticenergyacquiredbyit}\text{.}\\ {\text{Hence,kineticenergy,E}}_{\text{k}}\text{=}\frac{1}{2}{\text{mv}}^{\text{2}}\\ \therefore {\text{E}}_{\text{k}}\text{=}\frac{1}{2}\times 1500\times {\left(60\times \frac{5}{18}\right)}^{2}=208333.3{\text{}}^{\text{}}\text{J}\\ \text{Hence,20}{\text{8333.3}}^{\text{}}\text{Jofworkisrequiredtostopthecar}\text{.}\end{array}

Q:

An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

A:

\begin{array}{l}\text{Whenanobjectofmass}\left(\text{m}\right)\text{ismovingwithavelocity}\left(\text{v}\right)\text{then,itskineticenergy}\left(\text{K}\right)\text{isgivenas,}\\ \text{K=}\frac{1}{2}{\text{mv}}^{\text{2}}\\ \text{Inordertobringtheobjecttorest,anamountofworkisneedtobedonethat which is equaltothekineticenergy}\\ \text{oftheobject}\text{.}\\ \text{Hence,workdone}\left(\text{W}\right)\text{ontheobjecttobringittorest}\\ \text{wouldbe,W=}\frac{1}{2}{\text{mv}}^{\text{2}}.\end{array}

Q:

Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

A:

According to law of conservation of energy, energy can be neither created nor destroyed. It can only be converted from one form to another.

When a pendulum oscillates from its mean position O to either of its extreme positions P or Q, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it goes towards point O, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point O, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

After some time, bob comes to rest because air resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time.

Here, the law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Thus, the total energy of the pendulum and the surrounding system remain conserved.

Q:

An electric heater is rated 1500 W. How much energy does it use in 10 hours?

A:

\begin{array}{l}\text{Given},\text{Power,P=1500W=1}\text{.5kW}\\ \text{Time,t=10h}\\ \text{Now,Workdone=energyconsumedbytheheater}\\ \text{andEnergyconsumed=Power}\times \text{time}\\ \text{Hence, =1}\text{.5kW}\times \text{10h}\\ \text{=15kWh}\\ \text{Thus,energyconsumedbytheheaterin10his15kWh}\text{.}\end{array}

Q:

An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

A:

\begin{array}{l}\text{Given,Heightofobject,h=5m}\\ \text{massofobject,m=40kg}\\ \text{accelerationduetogravity,g=9}{\text{.8ms}}^{\text{-2}}\\ \text{Gravitationalpotentialenergyisgivenby,}\\ \text{W=mgh}\\ \text{or,=40kg\xd79}{\text{.8ms}}^{\text{-2}}\text{\xd75m=1960J}\\ \text{Now,potentialenergyathalf-way=}\frac{\text{1960J}}{\text{2}}\text{=980J}\text{.}\\ \text{Atthispoint,theobjecthasanequalamountofpotentialenergyandkineticenergy}\text{.}\\ \text{Thus,half-waydown,Kineticenergy=980J}\text{.}\end{array}

Q:

Certain force acting on 20 kg mass changes its velocity from 5 ms^{−1} to 2 ms^{−1}. Calculate the work done by the force.

A:

\begin{array}{l}\text{Kineticenergy}E\text{=}\frac{1}{2}m{v}^{2}\\ Where,\text{mass,m=20kg,v=velocity}\\ {\text{(i)KineticenergyE}}_{\text{1}}{\text{,whenvelocityis5ms}}^{\text{-1}}\\ {\text{K}}_{\text{1}}\text{=}\frac{1}{2}\left(20kg\right){\left(5m{s}^{-1}\right)}^{2}=250\text{J}\\ {\text{(ii)Kineticenergy(K}}_{2}{\text{)whenvelocityis2ms}}^{\text{-1}}\\ {\text{K}}_{2}=\frac{1}{2}\text{\hspace{0.17em}}\left(20kg\text{\hspace{0.17em}}\right)\text{\hspace{0.17em}}{\left(2\text{\hspace{0.17em}}m{s}^{-1}\right)}^{2}=\text{40J}\\ \text{Workdoneisequaltothechangeinthekineticenergy}\\ {\text{Hence,W=K}}_{2}-{\text{K}}_{\text{1}}\text{}=40-250=\text{-210J}\end{array}

Q:

Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.

• A donkey is carrying a load on its back.

• A wind-mill is lifting water from a well.

• A green plant is carrying out photosynthesis.

• An engine is pulling a train.

• Food grains are getting dried in the sun.

• A sailboat is moving due to wind energy.

A:

• While swimming, Suma exerts a force on the water in backward direction. Hence, she swims in the forward direction caused by the forward reaction of water. Thus, the force causes a displacement. Hence, Suma does the work while swimming.

• When the donkey carries a load, it applies a force in the upward direction while displacement of the load is in the forward direction. In this case, displacement is perpendicular to the force. Hence, the work done is zero.

• A wind mill pulls water against the gravitational force. Thus, work is done by the wind mill to lift the water from the well.

• In photosynthesis process, there is no displacement of the leaves of the plant. Hence, the work done is zero.

• A force is applied by engine on the train. Therefore, the train moves in the direction of applied force. Here, displacement is in the direction of force applied. Hence, work is done by the engine on the train.

• Food grains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in presence of the Sun.

• Wind exerts a force on the sailboat to push it in the forward direction. Thus, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

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