When a body falls freely towards the earth, then its total energy
A. increases
B. decreases
C. remains constant
D. first increases and then decreases
Since when a body falls the potential energy is converted to kinetic energy keeping the total energy conserved.
A car is accelerated on a labeled road and attains a velocity 8 times of its initial velocity. In this process the potential energy of the car
A. does not change
B. becomes twice to that of initial
C. becomes 4 times that of initial
D. becomes 16 times that of initial
P.E. = mgh, such that it does not depend on the velocity of the car.
In case of negative work the angle between the force and displacement is
A. 0°
B. 45°
C. 90°
D. 180°
Work done = force. Displacement.
Since the work is negative hence it means.
-W = F.s. cosθ
-W= Wcosθ
-1 = cosθ
θ = 180°
An iron sphere of mass 11 kg has the same diameter as an aluminum sphere of mass is 35 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
A. acceleration
B. momenta
C. potential energy
D. kinetic energy
Since KE, PE and momenta all depend on mass of the objects hence they will not be same.
A girl is carrying a school bag of 5 kg mass on her back and moves 300 m on a levelled road. The work done against the gravitational force will be (g =10 m s–2)
A. 6 ×103J
B. 6 J
C. 0.6 J
D. zero
Since she is walking on a leveled road hence the gravitational force will be constant.
Which one of the following is not the unit of energy?
A. joule
B. newton metre
C. kilowatt
D. kilowatt hour
Killo-watt is the unit of power.
The work done on an object does not depend upon the
A. displacement
B. force applied
C. angle between force and displacement
D. initial velocity of the object
W = F.s = Fs cosθ
It does not depend on the initial velocity of the object.
Water stored in a dam possesses
A. no energy
B. electrical energy
C. kinetic energy
D. potential energy
Water is stored at a height in damns such that it has potential energy in it. P.E. = mgh
A body is falling from a height h. After it has fallen a height h/2, it will possess.
A. only potential energy
B. only kinetic energy
C. half potential and half kinetic energy
D. more kinetic and less potential energy
Since PE is converted into KE, thus after covering a height of h/2 the ball’s PE will be converted to K.E.
A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly doubled, what will be the ratio of two kinetic energies?
Initial velocity = v =v’= 2v
Initial kinetic energy =
Final kinetic energy (K.E.)
(K.E) initial : (K.E) final = 1:4
Sunita can run with a speed of 8 m s–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 m s–1 against the frictional force of 25 N. Who is more powerful and why?
Power of Sunita = v.k
Where v = velocity and k = frictional constant
Power of Sunita = 10 N× 8 m/s= 80 W
Power of Kapil = 25 N× 3m/s = 75 W
So, Sunita is more powerful than Kapil.
A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round about (Fig. 11.1) of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.
k = 5 N, frictional force
Displacement according to diagram is:
1.5 km + 200 + 2km = 1500 + 200+ 2000 = 3,700 Km.
W = F.S
W = 5 × 3700 = 18,500 J.
Can any object have mechanical energy even if its momentum is zero? Explain.
Yes, Objects have mechanical energy even if its momentum is zero. Mechanical energy is sum of both potential energy and kinetic energy.
Momentum = mv = 0 ∴ v=o since m can never be zero.
Hence, there is no kinetic energy but the object may possess potential energy. Since PE is the product of mass, gravitational constant and height
Can any object have momentum even if its mechanical energy is zero? Explain.
No, any object cannot have momentum if its mechanical energy is zero. Mechanical energy is sum of both potential energy and kinetic energy.
Which means that there is no PE and KE.
= 0
⇒ v = 0
Now, momentum = mv = m(0) = 0
Hence, there will be no momentum.
The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given g = 10 m s–2)
Power = 2kW = 2000 W
Height = 10m
Power = work/ time = mgh/t ⇒ m (10)(10)/(60) = 2000
∴
The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A?
The PE will remain same on the earth and on planet A.
Mass is same on both the surface since we are talking about the same person, now weight = mg such that
Weight on earth = 2(weight on planet A)
Mg on earth = 2(Mg’ on planet A)
Equating PE on both the planets.
∴ Mgh on earth = Mg’h’ on planet A
2Mg’h = Mg’h’
2h =h’
h’ = 2(0.4) = 0.8 m
The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
To Prove: the increase in the kinetic energy of the body is equal to the work done by the force on the body i.e. W = (KE)f – (KE)i
Proof: According to equation of motion, we know that displacement and force can also be written as:
Now, W = F. ds
Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.
Yes, it is possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Since W = F.s where S is the displacement. So if the body is moving in a circular path then W=0.
A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2)
Height = 10m
Energy of the ball reduces by 40%
So, PE = m g h = m × 10 × 10 = 100m J.
Energy is reduced by 40% then the remaining energy is 100m – 40m = 60m J.
Now 60m = mx 10 x h’ ⇒ h’ = 6m
If an electric iron of 1200 W is used for 60 minutes everyday, find electric energy consumed in the month of April.
Power = 1200/1000 = 1.2 kW
Time – 60/60 = 1 hr
April has 30 days
E = power x time x days
= 1.2 x 1 x 30 = 36kWh
A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
Momentum = mv
Since both have same momentum, hence
P = p’
Mv = m’v’
If M > m
Then v < v’
Since, v <v’
hence KE < KE’
Thus, the light object has higher kinetic energy.
An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.
Mass of the car = 1000 Kg
Speed, V = 36 km/hr = (36 x 1000)/3600 = 10 m/s
Frictional force, k = 100 N
Power = k. V = 100 x 10 = 1000W
After collision of both the cars, according to law of conservation of momentum,
mAuA + mBuB = mAvA + mBvB
⇒ 1000 x 10 + 1000 x 0 = 1000 x 0 + 1000 x vB
⇒ vB = 10m/s
Therefore, the speed of the car (B) just after the collision is 10m/s.
A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m.
(a) How much work is done on the trolley?
(b) How much work is done by the girl?
Mass of the girl = 35 kg
Mass of trolley = 5kg
Initial velocity = 4m/s
Final velocity = 0m/s
Distance covered = 16m
Using the equation of motion,
Now Force = mass x acceleration,
(A) Work done on the trolley = 20 N ×16 m = 320 J
(B) Work done by the girl = 0 J.
Four men lift a 300 kg box to a height of 1 m and hold it without raising or lowering it.
(a) How much work is done by the men in lifting the box?
(b) How much work do they do in just holding it?
(c) Why do they get tired while holding it? (g = 10 m s–2)
(a) Force = mass x g = 300 x 10 = 3000N
Distance, s = 1m
W = F.s = 3000 J
(b) work done just holding it is zero; as the box does not move at all.
Such that W = F.s = F (0) = 0
(c) The force applied by the man in holding it is equal and opposite to the gravitational force applied on the box, while applying the force, muscular effort is involved. So they feel tired.
What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 40 m high. 2000 tones of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 m s–2)
Power is the rate of doing work.
Power = work/ time
Kilowatt is the unit of power and kilowatt hour is the unit of energy.
Height of the fall = 40 m,
mass of water falling= 2000 × 103 kg = 2 × 106 kg
Thus, the equivalent power is 1.25 x 107 W
How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 400 W, be able to lift at constant speed of 1 ms–1 vertically? (g = 10 ms–2)
Power = 400 W
Speed = 1m/s
g = 10 m/s2
Power = mgh/t = mgv
m = power/ g x speed = 400/ (10 x 1) = 40 Kg
Define watt. Express kilowatt in terms of joule per second. A 200 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?
One watt is the power of work done per unit time.
1 P = 1 J / 1 s .
1 kilowatt = 1000 J s–1
For 1 Kg power developed = 500W
Total Power for 200 kg = 200 × 500 = 1 × 104 W
Velocity of the car = 20m/s
Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 m s–2)
(i) a butterfly of mass 1.0 g that flies upward at a rate of 0.5 m s–1.
(ii) a 250 g squirrel climbing up on a tree at a rate of 0.5 m s–1.
(i) Mass of butterfly = 1 g = 1/1000
velocity= 0.5 m/s
Power = mg × velocity, g = 10 m s–2
(ii) Mass of squirrel = 250 g = 250/1000
velocity= 0.5 m/s
Power = mg × velocity, g = 10 m s–2
Hence, the power with which the squirrel is climbing is much higher than that of a butterfly flying.