In physics, work is said to be done only if the point of application of the force moves. If a person lifts the stone from the ground, some work is done because it has been displaced from the ground. The rate of doing work is called power. It is measured by the amount of work done in one second. Energy is the capacity of the body to do work.
Download JEE Advanced Previous Year Questions with Solutions on Work, Power and Energy PDF
Question 1) If the heart pushes 1 cc of blood in 1 s under pressure 20000 Nm-2, the power of the heart is
1) 0.02 W
2) 400 W
3) 5 × 10-10 W
4) 0.2 W
Answer: 1) 0.02 W
Solution:
Given,
Pressure, P = 20000 Nm-2
Volume per second = 1cc
Power of heart = Pressure x volume per second
= 20000 Nm-2 x 10-6 = 0.02 w
Question 2) A ball is dropped from a height of 20 m. If the coefficient of restitution is 0.9, what will be the height attained after the first bounce?
1) 1.62 m
2) 16.2 m
3) 18 m
4) 14 m
Answer: 2) 16.2 m
Solution:
Ball is dropped from a height, h= 20 m
coefficient of restitution = 0.9
Height after the first collision, h’ = e2h
= (0.9)2 x 20
= 16.2 m
Question 3) A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time, the particle goes up is
1) –0.5 J
2) –1.25 J
3) 1.25 J
4) 0.5 J
Answer: 2) –1.25 J
Solution:
Mass = 100 g
Speed, u = 5 m/s
Work done by the force of gravity is the kinetic energy while moving up
Wg = (1/2)mvf2 – (1/2)mvi2
= 0 – (1/2) x (0.1 kg) x 25
= -1.25 J
Question 4) A cubical vessel of height 1 m is full of water. What is the amount of work done in pumping water out of the vessel? (take, g =10 ms-2)
1) 1250 J
2) 5000 J
3) 1000 J
4) 2500 J
Answer: 2) 5000 J
Solution:
Mass of water in the cubical vessel= density x volume
m = v x ρ
= l3 x ρ
= 1 x 1000
= 1000 kg
Centre of mass of the cubical vessel, h = l/2 = 1/2
W = mgh
= 1000 x 10 x (1/2)
= 5000 J
Question 5) Potential energy of a particle is related to x coordinate by equation x2 – 2x will be equilibrium at
1) x = 0.5
2) x = 2
3) x = 1
4) x = 4
Answer: 3) x = 1
Solution:
For stable equilibrium, Fnet = 0
F = – dU/dx
F = -d(x2 – 2x)/dx
= -(2x -2)
-2x + 2 = 0
⇒ x = 2/2 = 1
⇒ x = 1
Question 6) A 16 kg block moving on a frictionless horizontal surface with a velocity of 4 m/s compresses an ideal spring and comes to rest. If the force constant of the spring be 100 N/m, then the spring is compressed by
1) 1.6 m
2) 4 m
3) 6.1 m
4) 3.2 m
Answer: 1) 1.6 m
Solution:
Kinetic energy of the block, K = (1/2)mv2
The kinetic energy will be equal to the work done by the block to compress the spring before coming to rest.
W = (1/2) kx2
Therefore,
(1/2)mv2 = (1/2) kx2
x = v√(m/k)
= 4 x √(16/100)
= 1.6 m
Question 7) A bomb of mass 9 kg explodes into two parts. One part of mass 3 kg moves with velocity 16 m/s, then the kinetic energy of the other part is
1) 162 J
2) 150 J
3) 192 J
4) 200 J
Answer: 3) 192 J
Solution:
Given,
Mass of the bomb = 9 Kg
Mass of smaller part = 3 kg
Mass of the bigger part = 6 Kg
Velocity, v = 16 m/s
Momentum before explosion = momentum after explosion
M x 0 = 3 x 16 + 6 x v
6v = – (3 x 16)
v = – 8 m/s
Kinetic energy of mass 6 Kg = (1/2) x 6 x (8)2
= 192 J
Question 8) The kinetic energy of a body becomes four times its initial value. The new momentum will be
1) same as the initial value
2) twice the initial value
3) thrice the initial value
4) half of its initial value
Answer: 2) twice the initial value
Solution:
E = p2/2m
E ∝ p2
(E1/E2) = p,2/p22
⇒ 1/4 = p,2/p22
⇒ 1/2 = p1/p2
p2 = 2p1
Question 9) A ball of mass 2 kg moving with velocity 3 ms-1, collides with spring of natural length 2m and force constant 144 Nm-1. What will be the length of the compressed spring?
1) 2 m
2) 1.5 m
3) 1 m
4) 0.5 m
Answer: 2) 1.5 m
Solution:
The kinetic energy of the ball = Potential energy of the spring
(1/2)mv2 = (1/2)kx2
(1/2) x 2 x (3)2 = (1/2) 144 x2
9 = 72x2
x = √9/72
x = 1/2√2 m
The length of the compressed spring
= [2 – 1/2√2]
= [4√2 – 1]/2√2
= 1.5 m
Question 10) A steel ball of mass 5 g is thrown downward with a velocity of 10 ms-1 from a height of 19.5 m and it penetrates sand by 50 cm. The change in mechanical energy will be (take, g =10ms-2 )
1) 1 J
2) 1.25 J
3) 1.5 J
4) 1.75 J
Answer: 2) 1.25 J
Solution:
Mass, m = 5g = 0.005 Kg
h = 19.5 m
x = 50 cm = 0.5 m
v = 10 m/s
The change in the mechanical energy, ΔU = mg(h+x) + (1/2)mv2
= 0.005 x 10(19.5 + 0.5) + (1/2)x 0.005 x (10)2
= 0.005 x 10 x 20 + (1/2)x 0.005 x 100
= 1 + 0.25 = 1.25 J
Also Read:
JEE Main Previous Year Solved Questions on Work, Power and Energy
Work, Power & Energy – Solved Questions
Work Power and Energy – Important Topics
Comments