JEE Advanced (Single Correct Type): Work, Power & Energy | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O given by F = -k/x²i . If the particle starts from rest at a distance a from the origin, the speed it will attain to reach a distance x from origin (a > x):
(a) 
(b) 
(c) 
(d) 

Correct Answer is option (a)

F = -k/x²

Q.2. A block of mass m is moving with a constant acceleration ‘a’ on a rough horizontal plane. If the coefficient of friction between the block and ground is μ, the power delivered by the external agent after a time t from the beginning is equal to:
(a) ma²t
(b) μmgat
(c) μm(a+μg)gt
(d) m(a+μg)at

Correct Answer is option (d)

Instantaneous power delivered = P = = Fv
F – f = ma
⇒ F = f + ma
⇒ P = (f + ma) v
Put f = μmg
∴ P = (μmg + ma)v = m(a + μg).at
Hence, (D) is correct.

Q.3. A spring placed horizontally on a rough horizontal surface is compressed against a block of mass m placed on the surface so as to store maximum energy in the spring.  If the coefficient of friction between the block and the surface is m , the potential energy stored in the spring is  
(a) 
(b) 
(c) 
(d) 

Correct Answer is option (a)

For equilibrium of the block,
F_max-mmg = 0
⇒ F = μmg
⇒ μmg = kx
⇒ x = μmg/k

Hence, (A) is correct.

Q.4. To one end of a string of length l a particle of mass m is attached, whereas the other end is fixed. A nail is located at a distance d below the point of suspension. For the ball to completely swing around in a circle centered on the nail, what is the value of d in terms of length l?
(a) 0.4 l
(b) 0.5 l
(c) 0.6 l
(d) 0.7 l

Correct Answer is option (c)
Let R be the radius of the circle. The particle velocity at B should be √5gR to complete the circle.

But d + R = l
⇒ d = 0.6 l
Hence, (C) is correct.

Q.5. A block of mass m moving with speed v compresses a spring through a distance x before its speed is halved. What is the value of spring constant?
(a) 3mv²/4x²
(b) mv²/4x²
(c) mv²/2x²
(d) 3mv²/x²

Correct Answer is option (a)
Initial kinetic energy =1/2 mv²
Final energy = 1/2(v/2)² + 1/2 kx²
By principle of conservation of energy,
1/2 mv² = 1/2 m v²/4 + 1/2 kx²

∴ k = 3mv²/4x²

Q.6. Small block of mass 100 g is pressed against a horizontal spring fixed at one end and compression is 5 cm. The spring constant is 100 N/m. When the block moves horizontally it leaves the spring. Where will it hit the ground 2 m below the spring?

(a) Horizontal distance of 1 m from end of spring.
(b) Horizontal distance of 2 m from end of spring.
(c) 0.5 m from free end of spring.
(d) 1.5 m from free end of spring.

Correct Answer is option (a)

Let v be the velocity when it leaves the spring.

Then 1/2 mv² = 1/2 kx².

v = 10 x 0.05 √10
Time to fall vertical distance of 2 metres from the spring

Horizontal distance  = 2/√10 x v  = 2/√10 x 10 x  0.05 x √10 = 1 m.

Q.7. A spring of natural length l and spring constant k is fixed on the ground and the other is fitted with a smooth ring of mass m which slides on a horizontal rod fixed at a height also equal to l (see Figure). Initially the spring makes an angle of 37° with the vertical when the system is released from rest. What is the speed of the ring when the spring becomes vertical?
(a) 
(b) 
(c)
(d)

Correct Answer is option (a)

In the initial position of the ring as shown in the Figure,

The length of spring = l/cos37°

∴ Extension =
= l/4

Energy stored in spring =
This stored energy when released becomes kinetic energy of the ring. If v is the velocity of the ring, kinetic energy when it is vertical = 1/2 mv²
By principle of conservation of energy,
 

Q.8. An ideal massless spring S can be compressed one metre by a force of 100 newton. The same spring is placed at the bottom of a frictionless inclined plane inclined at 30° to horizontal. A block M of mass 10Kg is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 metres. What is the speed of mass just before it reaches the spring? (g = 10 m/s²).
(a) √20 m/s
(b) √30 m/s
(c) √10 m/s
(d) √40m/s

Correct Answer is option (a)
Applied force on the spring, F = k.x,
where x is the distance through which it is compressed.
k = F/x = 100N/1m = 100 N/m
Let the mass M slide a distance s metres along the incline before hitting the spring. The spring gets compressed by 2 metres. Hence the mass M slides a total distance (s + 2) metres along the incline. Initially M is placed at a height (s + 2) sin 30° above the bottom of incline = s + 2/2 = h.

The mass has initially P.E. = Mgh = mg(s + 2)/2
When the spring is compressed, the energy has gone entirely into deformation of spring.

s = 40/2 - 2 = 2 m. The mass falls through a height, s sin 30° = s/2 .

Gain in K.E. = Loss of P.E.
1/2 Mv² = Mghor  v = √2gh

Q.9. A stone is projected at time t = 0 with a speed V₀ and an angle θ with the horizontal in a uniform gravitational field. The rate of work done (P) by the gravitational force plotted against time (t) will be as
(a)
(b)
(c)
(d)

Correct Answer is option (d)
Rate of work done is the power associated with the force. It means rate of work done by the gravitational force is the power associated with the gravitational force. Gravitational force acting on the block is equal to its weight mg which acts vertically downwards.

Velocity of the particle (at time t) has two components,
(i) a horizontal component v₀ cos θ, and
(ii) a vertically upward component (v₀ sin θ-gt)
Hence, the power associated with the weight mg will be equal to

This shows that the curve between power and time will be a straight line having positive slope but negative intercept on Y-axis.

Q.10. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
(a) 
(b) 
(c) 
(d) 

Correct Answer is option (d)

v_f can be found by C.O.E

Q.11. A particle moves along the x-axis from x = 0 to x = 5 m under the influence of a force given by F = (7 - 2x + 3x²) N.  The work done in the process is
(a) 70 J
(b) 270 J
(c) 35 J
(d) 135 J

Correct Answer is option (d)
W = ∫Fdx =
= (35 - 25 + 125) J = 135 J

Q.12. A particle starts from rest under gravity.  Its potential energy with respect to ground (PE) and its kinetic energy (KE) are plotted with time t.  Choose the correct graph.
(a) 
(b)
(c)
(d)

Correct Answer is option (b)
K.E. =
⇒ Parabolic
P.E. = mgh = mg(h -ut- 1/2gt²) =
⇒ Inverted parabola
⇒ B is correct.

Q.13 If a man increases his speed by 2 m/s, his K.E. is doubled. The original speed of the man is
(a) (2 + √2) m/s
(b) (2 + 2√2) m/s
(c) 4 m/s
(d) (1 + 2√2) m/s

Correct Answer is option (b)

Solve for v

Q.14. The kinetic energy K of a particle moving along a circle of radius R depends on the distance covered ‘s’ as K = as², where ‘a’ is a constant. The force acting on the particle is
(a) 2as²/R

(b) 

(c) 2as
(d) 2a

Correct Answer is option (b)

1/2 mV² = aS² ⇒ a_c = V²/R = 2aS²/mR . . . (1)

Q.15. The displacement of a particle of mass 1 kg on a horizontal smooth surface is a function of time given by x = 1/3 t³.  The work done by an external agent for first one sec is
(a) 0.5 J
(b) 2 J
(c) 0.60 J
(d) none of these

Correct Answer is option (a)
Acceleration =
∴ Force = 2t ∴ W = ∫Fdx =  = 0.5 J

Q.16. An object is acted upon by the forces N. If the displacement of the object is m, the kinetic energy of the object
(a) remains constant
(b) increases by 1 J
(c) decreases  by 1 J
(d) decreases by 2 J

Correct Answer is option (c)
W == (5 -6)J = -1 J

Q.17. A 500 kg elevator cab is descending with speed v₀ = 4 m/s when the winch system that lowers it begins to slip, allowing it to fall with a constant acceleration a = g/5. During its fall through a distance of 12 m, the total work done on the elevator cab is (Take g = 10 m/s²)
(a) 60 kJ
(b) -40 kJ
(c) 12 kJ
(d) 100 kJ

Correct Answer is option (c)
Work done by all forces = change in KE
= 12 kJ 

Q.18. A block of mass m, resting on a frictionless surface starts moving under the influence of a constant force F. The ratio of instantaneous power to average power developed by force F in time t is
(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 1 : 1

Correct Answer is option (b)
Instantaneous power, P₁ =
and average power, P₂ =
∴ P₁/P₂ = 2:1

Q.19. An engine develops 10 kW of power. How much time will it take to lift a mass of 250 kg to a height of 40 m?
(a) 10 sec
(b) 0.1 sec
(c) 5 sec
(d) 0.2 sec

Correct Answer is option (a)
Pt = mgh
t =

Q.20. A block of mass 2 kg was moving along a straight line on a smooth surface with a speed of 5 m/s. At t = 0, a force given by F = (3 + 2t) N directed in the direction of motion of the body starts acting on the block. The kinetic energy of the block after 2 sec is
(a) 20 J
(b) 200 J
(c) 100 J
(d) none of these

Correct Answer is option (c)

P = 20 kg- m/sec
 = 100 J