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.
Q2: Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v_{1} while body B is at rest before collision. The velocity of the system after collision is v_{2}. The ratio v_{1 }: v_{2} is: [2024]
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
Ans: (b)
In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body A is moving with a velocity v_{1} and body B is at rest. The conservation of momentum must hold true because no external forces are acting on the system.
The momentum before the collision is only due to body A since body B is at rest. Therefore, the total initial momentum p
where:
v_{0} = 0 is the velocity of body
Since they undergo a completely inelastic collision, bodies A and B stick together after the collision and hence move with a common velocity v_{2}. The total mass of the combined system postcollision is The momentum after the collision is given by:
Applying the conservation of momentum (since no external force implies momentum is conserved), we equate p
Dividing through by
Thus, solving for the ratio v_{1} / v_{2} gives:
Therefore, the ratio of v_{1} to v_{2} is
Option B: 2 : 1
Q2: A particle moves with a velocity horizontally under the action of constant force . The instantaneous power supplied to the particle is : [2023]
(a) 200 W
(b) Zero
(c) 100 W
(d) 140 W
Ans: (d)
Q1: The restoring force of a spring with a block attached to the free end of the spring is represented by [2022]
(a)
(b)
(c)
(d)
Ans: (a)
Where F is restoring force
x is displacement of block from equilibrium position
Q2: 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.
Q3: The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is [2022]
(a) 36
(b) 36
(c) 36
(d) 1
Ans: (a)
Energy = Power
E = 100
= 36
Q1: 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
Q2: 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
Q1: 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
Q2: 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
Q3: 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
Q4: 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})
Q1: 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
Q2: 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
Q3: 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
Q1: 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
Q2: 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
Q1: A particle moves from a point when a force of N is applied. How much work has been done by the force ? [2016]
(a) 8 J
(b) 11 J
(c) 5 J
(d) 2 J
Ans: (c)
Q2: 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
Q3: A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^{−4} J by the end of the second revoluation after the beginning of the motion ? [2016]
(a) 0.18 m/s^{2}
(b) 0.2 m/s^{2}
(c) 0.1 m/s^{2}
(d) 0.15 m/s^{2}
Ans: (c)
Given: Mass of particle, M = 10g
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10^{–4}J
acceleration a_{t }= ?
Now, using
Q4: What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
Q1: Two particles A and B, move with constant velocities . At the initial moment their position vectors are respectively. The condition for particles A and B for their collision is [2015]
(a)
(b)
(c)
(d) Ans: (c)
Q2: A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_{0}. It collides with the ground, losses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v_{0} is (Take g = 10 m s^{−2}) [2015]
(a) 28 m s^{−1}
(b) 10 m s^{−1}
(c) 14 m s^{−1}
(d) 20 m s^{−1}
Ans: (d)
When ball collides with the ground it loses its 50% of energy
Q3: The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6 × 103 kg/m3 and g = 10 m/s2 then the power (in watt) is [2015]
(a) 3.0
(b) 1.50
(c) 1.70
(d) 2.35
Ans: (c)
Here, Volume of blood pumped by man’s heart,
V = 5 litres = 5 × 10^{–3} m^{3} (
Time in which this volume of blood pumps, t = 1 min = 60 s
Pressure at which the blood pumps,
P = 150 mm of Hg = 0.15 m of Hg
= (0.15 m)(13.6 × 10^{3} kg/m^{3})(10 m/s^{2})
= 20.4 × 103 N/m^{2}
Q4: 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
Q5: 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
Q6: 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 byThe 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
102 videos411 docs121 tests

1. What is work in the context of physics? 
2. How is energy related to work? 
3. What is power in the context of physics? 
4. How can the concept of work, energy, and power be applied in reallife situations? 
5. How can one increase the power output in a mechanical system? 

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