1.Define work and energy.
3. List two conditions which need to be satisfied for the work to be done on an object.
Ans. Work is equal to the product of the force and displacement. Work done = force × displacement W = F x s.There for two conditions are there must be a force and displacement in the direction of force.
4.Work and energy are related to each other. Explain the given statement.
5. When is work done by a force zero? Is work have a direction?
6. Define 1 joule of work.
7. Write the expression for work.
20. What is the work done against gravity when a body is moved horizontally along a friction less surface?
23.What is the form of energy possessed by a running car?
31. Find the expression for gravitational potential energy of a body of mass m at a height h.
Ans. When a force acts on an object and the objects moves, we say the force has done work.
The capacity of an object to do work is called the energy of the object.
2.When do we say that work is done?
Ans. When a force acts on an object and the objects moves, we say the force has done work.
3. List two conditions which need to be satisfied for the work to be done on an object.
Ans. Work is equal to the product of the force and displacement. Work done = force × displacement W = F x s.There for two conditions are there must be a force and displacement in the direction of force.
4.Work and energy are related to each other. Explain the given statement.
Ans.Whenever work is done on an object, its energy increases. when we push an object it starts moving. We do work on the object and object acquires kinetic energy.
5. When is work done by a force zero? Is work have a direction?
Ans. If the displacement of of an object is perpendicular to the force acting on it, the work done by the force on the object is zero.
6. Define 1 joule of work.
Ans. When a force of 1N acts on an object and the object moves a distance 1 m in the direction of force, the work done by force is 1 J.
7. Write the expression for work.
Ans. Work is equal to the product of the force and displacement. Work done = force × displacement W = F x s.
8.When displacement of an object is perpendicular to the force acting on it. what is the work done by the force?
Ans. When displacement of an object is perpendicular to the force acting on it, the work done by the force is zero.
9. Write the expression for the work done by the force when displacement makes an angle with force.
Ans. The expression for the work done by the force when displacement makes an angle φ with force : W = f d cos φ
Where W = Work done, f = force, d = displacement and φ = angle between force and displacement.
10. Define kinetic energy.Write the expression of kinetic energy.
Ans. Energy possessed by an object due to its motion is called kinetic energy. The kinetic energy of an object increases with its speed.
K.E = 1/2 mv^2.
11.Define potential energy with expression.
Ans. The energy of an object because of its position or configuration or shape is called its potential energy.
P.E = m g h
12.What is mechanical energy?
Ans. The sum of potential energy and kinetic energy is called mechanical energy.
M.E = P.E + K.E
13. What is elastic potential energy How many watt in 1 horsepower?
Ans. The energy stored in the bow due to the change of shape or the energy acquired by the band in its stretched position is called elastic potential energy.
1 horsepower = 746 W.
14. What is the commercial unit of energy? State value of commercial unit of electrical energy in Joules.
14. What is the commercial unit of energy? State value of commercial unit of electrical energy in Joules.
Ans. The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour. This is called commercial unit of energy.
1 kWh= 1kW x 1h
1 kWh = 1000w x 3600s
1 kWh = 3.6 x 10^6 J.
15. What is the principle of conservation of energy?
Ans. Energy can neither be created nor destroyed. The total energy before and after the transformation remains the same. Energy can only be converted from one form to another.
16.Does work have a direction?
Ans. Work has only magnitude and no direction.
17. What happens to the potential energy of a body when its height is doubled?
Ans. The potential energy of a body will be doubled when its height is doubled because potential energy is directly proportional to height,as P.E = mgh.
18. If the angle between displacement and force acting on a particle is ∅. For which value of ∅ is the work done minimum and maximum?
Ans. When ∅ is 0 degree the work done is maximum and When ∅ is 90 degree the work done is minimum.
19. Moon is experiencing a gravitational force due to earth and is revolving around the earth in a circular orbit. How much work is done by moon?
Ans. Work done is zero because angle between gravitational force and displacement is 90 degree.
20. What is the work done against gravity when a body is moved horizontally along a friction less surface?
Ans. Zero because angle between gravitational force and displacement is 90 degree.
21. If the speed of a body is halved,what is the change in its kinetic energy?
Ans. Kinetic energy K.E = 1/2 mv^2
When the speed of a body is halved that is v/2
K.E' = 1/2 x m x (v/2)^2 = 1/2 x m x v^2/4 = 1/4 (1/2mv^) = 1/4 K.E
Kinetic energy becomes one forth.
22. By what factor does the kinetic energy of a particle increase if the speed is increased by a factor of 3?
Ans. The kinetic energy of a particle increase 9th times.
K.E = 1/2 mv^2
K.E' = 1/2 x m x (3v)^2
K.E' = 1/2 x m x (v)^2 x 9
K.E' = 9 (1/2 mv^2)
23.What is the form of energy possessed by a running car?
Ans. Kinetic energy.
24. A 2 m high person is holding a 25 kg trunk on his head and is standing at a roadway bus terminus. How much work is done?
Ans. Here person exert a vertically upward force on the trunk and the trunk has not moved any distance in the vertical direction, hence work done by person is zero.
25. A horse of mass 210 kg and a dog of mass 25 kg are running at the same speed. Which of the two possesses more kinetic energy?
Ans. The horse possessed more kinetic energy because kinetic energy is directly proportional to mass.
26. When displacement is in a direction opposite to the direction of force applied, what is the type of work done?
Ans.When displacement is in a direction opposite to the direction of force applied, work done is negative. W = (-d) x f = - df.
27. If the heart works 60 Joule in one minute,what is its power?
Ans. Given:
W = 60 J, t = 1 min. = 60 s.
Therefore P = W/t = 60/60 = 1w.
A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle ?
28. Define power and 1 watt of power .Write the S.I unit of power.
Ans.The rate of doing work or the rate of transfer of energy is called power. If an agent does a work W in time t, then power is given by:
Power = work/time or P = W/t
The unit of power is watt.
29.The potential energy of a free falling body decreases progressively. Does this violate the law of conservation of energy? Why?
Ans. During the free fall of the object, the decrease in potential energy, at any point in its path, appears as an equal amount of increase in kinetic energy. There is thus a continual transformation of gravitational potential energy into kinetic energy. So no the law of conservation of energy violated.
30. Calculate the work done when a force of 15 N moves a body by 5 m in its direction.
Ans. Work done W = F x D
Given F = 15 N ans D = 5 m.
W = 15 N x 5m = 75J
31. Find the expression for gravitational potential energy of a body of mass m at a height h.
Ans. Consider an object of mass, m. Let it be raised through a height, h from the ground. The minimum force required to raise the object is equal to the weight of the object, mg. The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done, W = force × displacement = mg × h = mgh .Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy of the object.
Ep = mgh.
32. Find the expression for kinetic energy energy of a body of mass m moving at a speed v.
Ans. Consider an object of mass, m moving with a uniform velocity, u. Let it now be displaced through a distance s when a constant force, F acts on it in the direction of its displacement.
The work done, W = F x s. The work done on the object will cause a change in its velocity. Let its velocity change from u to v. Let a be the acceleration produced.
From the equation of motion
v^2 – u^2 = 2as
This gives s = v^2 – u^2/2a
We know F = m a.
We can write W = ma x v^2 – u^2/2a = 1/2 m (v^2 – u^2)
If the object is starting from its stationary position, that is, u = 0
then W = 1/2 mv^2
Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity v K.E = 1/2 mv^2.
33. A body is thrown vertically upwards. Its velocity goes on decreasing. Write the change in kinetic energy when its velocity becomes zero.
Ans. Kinesthetic energy becomes zero, because K.E = 1/2 mv^2
K.E = 1/2 x m x 0 = 0.
34. A car and a truck are moving with the same velocity of 60 km/hr. Who of one has more kinetic energy?
Ans. Truck has more kinetic energy because mass of truck is more than car and kinetic energy directly proportional to mass.
35. What will be the kinetic energy of a body when its mass is made four times and velocity is double?
Ans. If the mass of body be = m kg
velocity = v m/s
then Kinetic energy = 1/2mv^2
When its mass is made four times and velocity is double that means :
m becomes 4m and v becomes 2v,
Kinetic energy will be = 1/2 x 4m x (2v)^2
= 1/2 x 4m x 4v^2
= 16 (1/2mv^2)
Kinetic energy becomes 16 times the initial value.
36. A light and heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has larger kinetic energy?
Ans. Let the mass and velocity of light object be m and v.
And the mass and velocity of heavy object be M and V.
Therefore mv=MV
& m/M = V/v
The ratio of their kinetic energies= 1/2 mv^2
1/2 MV^2
= m/M(M/m)^2
M/m
=mass of heavier object/mass of lighter object
The heavy object has more K.E because K.E is directly proportional to mass.
K.E is directly proportional to the mass.
The heavy object has more mass.So,it has more kinetic energy.
37. A freely falling body eventually stops on the ground. What happens to its kinetic energy on the ground?
Ans. The kinetic energy becomes zero because there is no velocity but changes into heat and sound due to collision between object and ground.
38. A rocket is moving up with a velocity v if the velocity of this rocket is suddenly tripled,what will be the ratio of two kinetic energies?
Ans. When a rocket of mass m is moving with velocity v its K.E will be 1/mv^2, and when the velocity of this rocket is suddenly tripled,the ratio of two kinetic energies will be = 1/2 mv^2 : 1/2 m (3v)^2
= v^2/9v^2
= 1:9 Ans.
39. The kinetic energy of an object of mass 'm' moving with a velocity of 5 m/s is 25 J. Calculate its kinetic energy when its velocity is double.
Ans. In first case:
K.E = 1/2 mv^2
25 = 1/2 x m x 5^2
25 = 1/2 x m x 25
m = 25 x 2
25
m= 2kg
Now when its velocity is double,
then K.E = 1/2 mv^2
K.E = 1/2 x 2 x (2 x 5)^2
= 100 J Ans.
40. Write the form of energy possessed by the object in the following situation :
(i) a coconut falling from the tree .
1. At what speed a body of mass 1 kg will have a kinetic energy of 1 J?
2. What would be the amount of work is done on an object by a force, if the displacement of the object is zero?
3. How much work is done by a weight lifter when he hold a weight of 80 kg on his shoulder's for two minutes ?
Ans. Truck has more kinetic energy because mass of truck is more than car and kinetic energy directly proportional to mass.
35. What will be the kinetic energy of a body when its mass is made four times and velocity is double?
Ans. If the mass of body be = m kg
velocity = v m/s
then Kinetic energy = 1/2mv^2
When its mass is made four times and velocity is double that means :
m becomes 4m and v becomes 2v,
Kinetic energy will be = 1/2 x 4m x (2v)^2
= 1/2 x 4m x 4v^2
= 16 (1/2mv^2)
Kinetic energy becomes 16 times the initial value.
36. A light and heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has larger kinetic energy?
Ans. Let the mass and velocity of light object be m and v.
And the mass and velocity of heavy object be M and V.
Therefore mv=MV
& m/M = V/v
The ratio of their kinetic energies= 1/2 mv^2
1/2 MV^2
= m/M(M/m)^2
M/m
=mass of heavier object/mass of lighter object
The heavy object has more K.E because K.E is directly proportional to mass.
K.E is directly proportional to the mass.
The heavy object has more mass.So,it has more kinetic energy.
37. A freely falling body eventually stops on the ground. What happens to its kinetic energy on the ground?
Ans. The kinetic energy becomes zero because there is no velocity but changes into heat and sound due to collision between object and ground.
38. A rocket is moving up with a velocity v if the velocity of this rocket is suddenly tripled,what will be the ratio of two kinetic energies?
Ans. When a rocket of mass m is moving with velocity v its K.E will be 1/mv^2, and when the velocity of this rocket is suddenly tripled,the ratio of two kinetic energies will be = 1/2 mv^2 : 1/2 m (3v)^2
= v^2/9v^2
= 1:9 Ans.
39. The kinetic energy of an object of mass 'm' moving with a velocity of 5 m/s is 25 J. Calculate its kinetic energy when its velocity is double.
Ans. In first case:
K.E = 1/2 mv^2
25 = 1/2 x m x 5^2
25 = 1/2 x m x 25
m = 25 x 2
25
m= 2kg
Now when its velocity is double,
then K.E = 1/2 mv^2
K.E = 1/2 x 2 x (2 x 5)^2
= 100 J Ans.
40. Write the form of energy possessed by the object in the following situation :
(i) a coconut falling from the tree .
(ii) an object raised to a certain height.
(iii) blowing wind.
(iv) a child driving a bicycle.
Ans. (i) a coconut falling from the tree --- K.E
41.Given below are a few situations,study them and state in which of the given cases work is said to be done. Give reasons for your answer.
(i) A person pushing hard a huge rock does not move
(ii) A bullock pulling a cart up to 1 km on road.
(iii) A girl pulling a trolley for about 2 m distance.
(iv) A person standing with a heavy bag on his head.
Ans. (i) A person pushing hard a huge rock does not move --No displacement , so no work is be done.
(ii) A bullock pulling a cart up to 1 km on road-- There is 1 km displacement, so work is done.
(iii) A girl pulling a trolley for about 2 m distance -- There is displacement so work is done.
(iv) A person standing with a heavy bag on his head-- There is no displacement so no work is done.
Do these yourself 👇Ans. (i) a coconut falling from the tree --- K.E
(ii) an object raised to a certain height --- P.E
(iii) blowing wind -- K . E
(iv) a child driving a bicycle -- K.E 41.Given below are a few situations,study them and state in which of the given cases work is said to be done. Give reasons for your answer.
(i) A person pushing hard a huge rock does not move
(ii) A bullock pulling a cart up to 1 km on road.
(iii) A girl pulling a trolley for about 2 m distance.
(iv) A person standing with a heavy bag on his head.
Ans. (i) A person pushing hard a huge rock does not move --No displacement , so no work is be done.
(ii) A bullock pulling a cart up to 1 km on road-- There is 1 km displacement, so work is done.
(iii) A girl pulling a trolley for about 2 m distance -- There is displacement so work is done.
(iv) A person standing with a heavy bag on his head-- There is no displacement so no work is done.
42. An object thrown at a certain angle from 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 same horizontal line. What is the work done by the force of gravity on the object?
Ans. The force of gravity on the object acts vertically downwards. The displacement of the object is in the horizontal direction, so no work is said to be done.
43. A porter lift a luggage of 20 kg from the ground and puts in on his head 1.7 m above the ground. Find the work done by the porter on the luggage?
Ans. When porter lift a luggage from the ground and puts it on the head, porter applied force on upward direction on the luggage and there is a displacement of 1.7m in upward direction can say that work to be done.
Here m = 20kg, h=1.7m and g=10m/s
Work done in upward direction against gravity = mgh
Work = 20 x 10 x 1.7
Work = 340 J .
44. An archer stretched a bow to release an arrow to hit the target at distance of 10 m. Explain who does the work. In which form is the energy possessed by the bow and the arrow?
Ans. Archer stretched the bow to change its shape, so archer does the work and the energy possessed by the bow is potential energy due to change in its position or shape and the energy possessed by the arrow is K.E due to moving to the target.
45. A body of mass 2 kg is thrown up with a speed of 25 m/s. Find the maximum potential energy.
Ans. Here mass of the body is =2 kg
Initial speed u = 25 m/s
Final speed v = zero (where object stop and start falling,here potential energy maximum).
h = v^2 - u^2 / 2g
h= 0 - 625 /2 x 10 ( g= 10 m/s)
Now Potential energy = mgh
P.E= 2 x 10 x 625 /20=
P.E = 625J.
46. A 5 kg ball is thrown upward with a speed of 10 m/s.
(i) Calculate the maximum height attained by it.
(ii) Find the potential energy when it reaches the highest point .
Ans. Here mass of the body is =5 kg
Initial speed u = 10 m/s
Final speed v = zero (where object stop and start falling,here potential energy maximum).
Let the max. height be = h
h = v^2 - u^2 / 2g [ 2gh = v^2 - u^2]
h= 0 - 100 /2 x 10 ( g= 10 m/s)
Now Potential energy = mgh
P.E= 5 x 10 x 100 /20=
P.E = 250 J.
47. What is the work to be done to increase the velocity of a car from 30 km /hr to 40 km/hr, if the mass of the car is 1500 kg ?
Ans.Here initial velocity u = 30 km /hr = 30 x 5/18 = 8.33 m/s
Final velocity v = 40 km /hr = 40 x 5/18 = 16.67 m/s
mass m = 1500 kg.
W = F x S
W = m x a x v^2-u^2 = m x a x (v+u) (v-u)
2a 2a
[ F = ma and 2as = v^2 - u^2]
W = 1500 x (16.67 + 8.33) (16.67 - 8.33)/2
W = (1500 x 25 x 8.34)/2
W = 312750 /2
W = 156,375 J.
43. A porter lift a luggage of 20 kg from the ground and puts in on his head 1.7 m above the ground. Find the work done by the porter on the luggage?
Ans. When porter lift a luggage from the ground and puts it on the head, porter applied force on upward direction on the luggage and there is a displacement of 1.7m in upward direction can say that work to be done.
Here m = 20kg, h=1.7m and g=10m/s
Work done in upward direction against gravity = mgh
Work = 20 x 10 x 1.7
Work = 340 J .
44. An archer stretched a bow to release an arrow to hit the target at distance of 10 m. Explain who does the work. In which form is the energy possessed by the bow and the arrow?
Ans. Archer stretched the bow to change its shape, so archer does the work and the energy possessed by the bow is potential energy due to change in its position or shape and the energy possessed by the arrow is K.E due to moving to the target.
45. A body of mass 2 kg is thrown up with a speed of 25 m/s. Find the maximum potential energy.
Ans. Here mass of the body is =2 kg
Initial speed u = 25 m/s
Final speed v = zero (where object stop and start falling,here potential energy maximum).
h = v^2 - u^2 / 2g
h= 0 - 625 /2 x 10 ( g= 10 m/s)
Now Potential energy = mgh
P.E= 2 x 10 x 625 /20=
P.E = 625J.
46. A 5 kg ball is thrown upward with a speed of 10 m/s.
(i) Calculate the maximum height attained by it.
(ii) Find the potential energy when it reaches the highest point .
Ans. Here mass of the body is =5 kg
Initial speed u = 10 m/s
Final speed v = zero (where object stop and start falling,here potential energy maximum).
Let the max. height be = h
h = v^2 - u^2 / 2g [ 2gh = v^2 - u^2]
h= 0 - 100 /2 x 10 ( g= 10 m/s)
Now Potential energy = mgh
P.E= 5 x 10 x 100 /20=
P.E = 250 J.
47. What is the work to be done to increase the velocity of a car from 30 km /hr to 40 km/hr, if the mass of the car is 1500 kg ?
Ans.Here initial velocity u = 30 km /hr = 30 x 5/18 = 8.33 m/s
Final velocity v = 40 km /hr = 40 x 5/18 = 16.67 m/s
mass m = 1500 kg.
W = F x S
W = m x a x v^2-u^2 = m x a x (v+u) (v-u)
2a 2a
[ F = ma and 2as = v^2 - u^2]
W = 1500 x (16.67 + 8.33) (16.67 - 8.33)/2
W = (1500 x 25 x 8.34)/2
W = 312750 /2
W = 156,375 J.
1. At what speed a body of mass 1 kg will have a kinetic energy of 1 J?
2. What would be the amount of work is done on an object by a force, if the displacement of the object is zero?
3. How much work is done by a weight lifter when he hold a weight of 80 kg on his shoulder's for two minutes ?
4. What is the work done by the earth in moving around the Sun?
5. Write the formula or expression for change in kinetic energy of a body of mass which experience a change in its velocity from u m/s to v m/s when force is applied on it.
6.A horse of mass 200 kg and a dog of mass 20 kg are running at the same speed. Who h of the two possesses more kinetic energy?
7.A box about 10 kg is placed at a point A on a horizontal surface. It is raised to a point B which is just at a distance of 2 m above A. Find the work done against by the force of gravity on the box. Justify your answer. ( g=10 m/s)
8. A bag of wheat 60 kg. Find the height to which it is lifted so that its potential energy is 3000 J. ( g= 10 m/s) .
9. When will be work positive,negative or zero ? Explain with examples.
10. Write the names of four different forms of energy and give an example where energy is transformed from one form to another?
9. When will be work positive,negative or zero ? Explain with examples.
10. Write the names of four different forms of energy and give an example where energy is transformed from one form to another?
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