1. When a body falls freely towards the earth, then its total energy
(c) remains constant
(d) first increases and then decreases
2. A car is accelerated on a leveled road and attains a velocity 4 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
3. In case of negative work the angle between the force and displacement in degree is
(a) 0 (b) 45 (c) 90 (d ) 180
4. An iron sphere of mass 10 kg has the same diameter as an aluminum sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(c) potential energy
(d) kinetic energy
5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 m s–2)
(a) 6 ×103 J
(b) 6 J
(c) 0.6 J
6. Which one of the following is not the unit of energy?
(b) newton metre
(d) kilowatt hour
7. The work done on an object does not depend upon the
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object
8. Water stored in a dam possesses
(a) No energy
(b) Electrical energy
(c) kinetic energy
(d) Potential energy
9. 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
10. How are Joule (J) and ergs (erg) related?
(a) 1J = 107 erg (b) 1erg = 10 -7J (c) 1J = 10-7 erg (d) None
1. (c) 2. (a) 3. (d) 4. (a) 5. (d) 6. (c) 7. (d) 8. (d) 9. (c) 10.(b)
10. 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: Initial velocity = u, then v = 3 u
Initial kinetic energy = 1/2 m u2
Final kinetic energy (K.E.) =1/2 m v′2 =1/2m (3u)2 =9 x (1/2 m u2 )
(K.E) initial : (K.E) final=1:9
11. Avinash 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?
Ans: Power of Avinash PA = FA x vA = 10 × 8 = 80 W
The power of Kapil Pk = Fk x vk = 25 × 3 = 75 W
So, Avinash is more powerful than Kapil.
12. 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 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.
Ans: F = 5 N
W = F.S = 5 × [1500 + 200 + 2000] = 18500 J.
13. Can any object have mechanical energy even if its momentum is zero? Explain.
Ans: Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.
14. Can any object have momentum even if its mechanical energy is zero? Explain.
Ans: No. Since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.
15. 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)
p=w/Dt=mgh/Dt = [(m x10 x10)/60]
m = 1200kg
Q16. If you push vigorously against a brick wall, how much work do you do on the wall? Explain.
Ans:There is no work done on the wall as there is no displacement of the wall.
Work = W = F S CosQ
F - Force applied S - Displacement Q - Angle between F and S
Work will be zero if F or S = 0 or angle Q = 90
In the current situation the wall is not moving and S is therefore zero. It follows that Work is zero, that is no work is done.
Q 17. 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. Find the change in momentum of the ball as a result of the collision with the side walk (in kg m/s).
Ans: Mass of the rubber ball = 0.2 kg = m
Initial velocity = 30 m/s = v1
Final velocity after collision = -20m/s = v2
Change in momentum = mv1 - mv2= m(v1 - v2)
= 0.2[30 - (-20)] = 10 kg m /s
Q.18. A 2.0 kg block is thrown upward from a point 20m above Earth's surface. At what height above Earth's surface will the gravitational potential energy of the Earth block system have increased by 500J?
Ans. Mass of the block m = 2.0 kg
P.E. of the block at a height of 20 m is = mgh = 2 x 9,8 x 20 = 392 J
Increase in P.E. = 500 J
Let the new height be h
New P.E. = mgh
mgh -392 = 500
mgh = 500 + 392 = 892 J
therefore h = 892/2x9.8 = 45.51m
Q.19. A person pushes a 10 kg cart a distance of 20 meters by exerting a 60 Newton horizontal force. The frictional resistance force is 50 Newtons. How much work is done by each force acting ont he cart? How much kinetic energy does the cart have at the end of the 20 meters if it started from rest:
Mass of cart = 10 kg Distance moved = 20 m
Horizontal force = 60 N Frictional force = 60 N
Frictional resistance force = 50 N Net force = 60 - 50 = 10 N
Work done by the horizontal force = 60 x 20 = 1200 J
Work done by the frictional force = - 50 x 20 = -1000 J
(It is negative because force and displacement are opposite to each other)
Net work done = 1200 - 1000 = 200 J = Change in the K.E. of the cart.
If the cart started from rest, the initial K.E. = 0
Final K.E. - Initial K.E. = 200 J
So Final K.E. = 200J
Q 20. Assuming an efficiency of 25% for the muscle system in the process of converting food energy into mechanical work, how much energy would be used by a person of mass 75 kg (weight 165lb) in the process of climbing four flights of stairs for a total height of 15 metes? Find the answer in joules and convert to dietary calories (1 dietary calorie = 4186 Joules)
Efficiency ? = 25%
Mass of the person = 75 kg
Height climbed = h = 15 m
Work done = gain in P.E. = mgh = 75 x 9.8 x 15 = 11025 J
Since the efficiency of the muscle system
= 25%, the food energy required to do this work
= (100 x 11025)/25 = 44100 J
4186 Joules = 1 dietary calorie
44100 Joules = 44100 / 4186 = 10.5 dietary calories.
The potential energy of a 0.2 kg particle moving along the x axis is given by:
U(x) = 8x2 - 2x4 where U is in Joules and x in meters.
When the particle is at x = 10 m, what will be its acceleration?
m = 0.2 kg
P.E. U(x) = 8x2 - 2 x4
here U is in Joules and x in meters
Intensity T = -du/dx
=- -d[8x2 - 2 x4]/dx = -(16x -8x)
The intensity is equal to the force per unit mass which is the same as acceleration
a = -(16x - 8x)|x = 10 m = -(160 -80) = -80 m/s2
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