The terms ‘work’, ‘energy’, and ‘power’ are frequently used in everyday language. In physics, however, the word ‘Work’ covers a definite and precise meaning. This chapter of NEET physics covers the fundamental concepts of energy, work, and power, and their applications to various physical systems. Let's have a look at the Previous Year's Questions of NEET Exam of this chapter.
Q.1. The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be [2023]
(a) 2 U
(b) 4 U
(c) 8 U
(d) 16 U
Ans: d
Q.2. A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is [2022]
(a) √2v
(b) 2√2v
(c) 3√2v
(d) v
Ans: b
Momentum of the system would remain conserved.
Initial momentum = 0
Final momentum should also be zero.
Let masses be 2m, 2m, and m
Q.3. An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms^{–1} . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s ^{–2} ) [2022]
(a) 20000
(b) 34500
(c) 23500
(d) 23000
Ans: b
The power delivered by the motor to the lift can be calculated using the formula:
Power = Force x Velocity
In this case, the force is the sum of the weight of the lift and the passengers and the frictional force opposing the motion:
Force = (Weight of lift + passengers) + Frictional force
Weight of lift + passengers = Mass x gravity
= (2000 kg / 1000) x 10 m/s^{2}
= 20,000 N
Therefore,
Force = (20,000 N) + (3000 N)
= 23,000 N
Now, using the formula for power, we get:
Power = Force x Velocity
= (23,000 N) x (1.5 m/s)
= 34,500 W
Therefore, the minimum power delivered by the motor to the lift in watts is 34,500 W.
Q.4. A particle is released from height S from the surface of the earth. At a certain height, its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively: [2021]
(a)
(b)
(c)
(d)
Ans: B
U + KE = E
4U = E = mgS
4mgh = mgS
h = S/4
Q.5. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?(g = 10 m/s^{2}) [2021]
(a) 12.3 kW
(b) 7.0 kW
(c) 10.2 kW
(d) 8.1 kW
Ans: D
Mass of water falling/second = 15 kg/s h = 60 m, g = 10 m/s^{2}, loss = 10 % i.e., 90% is used. Power generated = 15 x 10 x 60 x 0.9 = 8100 W = 8.1 kW.
Q.6. The energy required to break one bond in DNA is 10^{–20}J. This value in eV is nearly: [2020]
(a) 0.06
(b) 0.006
(c) 6
(d) 0.6
Ans: a
1.6 × 10^{–19} Joule = 1 eV
Q.7. Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is: [2019]
(a) 1/9
(b) 8/9
(c) 4/9
(d) 5/9
Ans: b
Fractional loss of KE of colliding body
Q.8. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it? [2019]
(a) 3 J
(b) 30 kJ
(c) 2 J
(d) 1 J
Ans: a
Work required = change in kinetic energy Final KE = 0
Initial KE
ΔKE = 3 J
Q.9. A force F = 20 + 10 y acts on a particle in y direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is [2019]
(a) 30 J
(b) 5 J
(c) 25 J
(d) 20 J
Ans: c
Work done by variable force is
Q.10. The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is: [2019]
(a) mgR
(b) 2mgR
(c)
(d)
Ans: c
Initial potential energy at earths surface is
Final potential energy at height h = R
As work done = Change in PE∴ W = U_{f} – U_{i}
(∵ GM = gR^{2})
Q.11. A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be : [2018]
(a) 0.5
(b) 0.25
(c) 0.8
(d) 0.4
Ans: b
Q.12. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to [2018]
(a) 3/2 D
(b) D
(c) 7/5 D
(d) 5/4 D
Ans: d
Q.13. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then [2018]
(a) KA < KB < KC
(b) KA > KB > KC
(c) KB < KA < KC
(d) KB > KA > KC
Ans: b
Q.14. A spring of force constant k is cut into lengths of ratio 1: 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is [2017]
(a) 1 : 6
(b) 1 : 9
(c) 1 : 11
(d) 1 : 14
Ans: c
Force constant of spring
after cutting the spring in the ratio 1:2:3, the force constants will be in the ratio 6:3:2
Force constant in series K' is given by
Force constant in parallel K" : 6k+3k+2k = 11k;
hence K'/K" = 1/11
Q.15. Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s^{2}. The work done by the [2017]
(i) gravitational force and the
(ii) resistive force of air is :
(a) (i) 1.25 J (ii) – 8.25 J
(b) (i) 100 J (ii) 8.75 J
(c) (i) 10 J (ii) – 8.75 J
(d) (i) – 10 J (ii) – 8.25 J
Ans: c
Q.16. A body of mass 1 kg begins to move under the action of a time dependent force , whenare unit vectors along x and y axis. What power will be developed by the force at the time t ? [2016]
(a) (2t^{3}+ 3t^{5})W
(b) (2t^{2} + 3t^{3})W
(c) (2t^{2} + 4t^{4})W
(d) (2t^{3} + 3t^{4})W
Ans: a
Q.17. Two particles of masses m_{1}, m_{2} move with initial velocities u_{1} and u_{2}. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v_{1} and v_{2} then we must have : [2015]
(a)
(b)
(c)
(d)
Ans: d
Q.18. Two similar springs P and Q have spring constants K_{P} and K_{Q}, such that K_{P} > K_{Q}. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_{P} and W_{Q} are related as, in case (a) and case (b), respectively: [2015]
A: W_{P} < W_{Q}; W_{Q} < W_{P}
B: W_{P} = W_{Q}; W_{P} > W_{Q}
C: W_{P} = W_{Q}; W_{P} = W_{Q}
D: W_{P} > W_{Q}; W_{Q} > W_{P}
Ans: d
Q.19. A mass m moves in a circle on a smooth horizontal plane with velocity v_{0} at a radius R_{0}. The mass is attached to a string which passes through a smooth hole in the plane as shown.
The tension in the string is increased gradually and finally m moves in a circle of radius R_{0}/2. The final value of the kinetic energy is : [2015]
(a)
(b)
(c)
(d)
Ans: d
When a mass moves in a circle of radius R_{0} with velocity v_{0}, its kinetic energy is given by
The tension in the string is gradually increased and the radius of the circle decreased to R_{0}/2. When the radius of the circle is R the tension in the string is the same as the centripetal force.
where L = mRv is the angular momentum which is conserved.
Work done in reducing the radius of the circle from R_{0} to R_{0}/2 is
The block of mass M = 10 kg is moving in the x  direction with a speed v = 10 m/s. Its initial kinetic energy is
It is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m.
Work done is given by
Final kinetic energy is, KE_{f} = KE_{i} + W = 500  25 = 475
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