JEE Advanced PYQ Work, Energy and Power – Energy

Q 1 :

A block $(B)$ is attached to two unstretched springs $S_{1}$ and $S_{2}$ with spring constants $k$ and $4 k$ , respectively (see fig. I). The other ends are attached to identical supports $M_{1}$ and $M_{2}$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block $B$ is displaced towards wall 1 by a small distance $x$ (figure II) and released. The block returns and moves a maximum distance $y$ towards wall 2. Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$ . The ratio $y / x$ is [2008]

[IMAGE 111]

4
2
1/2
1/4

1

2

3

4

(3)

Here when the block B is displaced towards wall 1, only spring $S_{1}$ is compressed and $S_{2}$ is in its natural state as the other end of $S_{2}$ is free.

Therefore the energy stored in the system $= \frac{1}{2} k_{1} x^{2}$

When the block is released, it will come back to the equilibrium position, gain momentum, overshoot the equilibrium position and move towards wall 2. As this happens, the spring $S_{1}$ comes to its natural length and $S_{2}$ gets compressed. The P.E. stored in the spring $S_{1}$ gets stored as the P.E. of spring $S_{2}$ when the block $B$ reaches its extreme position after compressing $S_{2}$ by $y$ . It is because no friction anywhere.

So, energy is conserved

$∴$ $\frac{1}{2} k_{1} x^{2} = \frac{1}{2} k_{2} y^{2} \Rightarrow \frac{1}{2} k x^{2} = \frac{1}{2} \times 4 k y^{2}$

$\Rightarrow x^{2} = 4 y^{2}$ $∴$ $\frac{y}{x} = \frac{1}{2}$

Q 2 :

A particle is acted by a force $F = k x$ , where $k$ is a +ve constant. Its potential energy at $x = 0$ is zero. Which curve correctly represents the variation of potential energy of the block with respect to $x$ [2004]

[IMAGE 112]
[IMAGE 113]
[IMAGE 114]
[IMAGE 115]

1

2

3

4

(2)

$For conservative forces Δ U = - W$

$Δ U = - \int_{0}^{x} F d x or Δ U = - \int_{0}^{x} k x d x$

$\Rightarrow U_{(x)} - U_{(0)} = - \frac{k x^{2}}{2} (∵ U_{(0)} = 0)$

$∴$ $U_{(x)} = - \frac{k x^{2}}{2} \Rightarrow x^{2} = \frac{- 2 U_{x}}{k}$

It represents a parabola below the $x$ -axis symmetrical.

Q 3 :

An ideal spring with spring-constant $k$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [2002]

$\frac{4 M g}{k}$
$\frac{2 M g}{k}$
$\frac{M g}{k}$
$\frac{M g}{2 k}$

1

2

3

4

(2)

Let $x$ be the maximum extension of the string. Here mechanical energy is conserved, so decrease in the gravitational potential energy of spring mass system $(M g x)$

= gain in spring elastic potential energy $(\frac{1}{2} k x^{2})$

$M g x = \frac{1}{2} k x^{2} \Rightarrow x = \frac{2 M g}{k}$

Q 4 :

A particle, which is constrained to move along the $x$ -axis, is subjected to a force in the same direction which varies with the distance $x$ of the particle from the origin as $F (x) = - k x + a x^{3} .$ Here $k$ and $a$ are positive constants. For $x \geq 0$ , the functional form of the potential energy $U (x)$ of the particle is [2002]

[IMAGE 116]
[IMAGE 117]
[IMAGE 118]
[IMAGE 119]

1

2

3

4

(4)

Q 5 :

Consider an elliptical shaped rail $P Q$ in the vertical plane with $O P = 3 m$ and $O Q = 4 m$ . A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is $(n \times 10)$ joules. The value of $n$ is (take acceleration due to gravity $= 10 {ms}^{- 2}$ ) [2014]

[IMAGE 120]

(5)

Work done = Increase in P.E. + gain in K.E.

$F \times d = m g h + gain in K.E.$

$18 \times 5 = 1 \times 10 \times 4 + gain in K.E.$

$∴$ $Gain in K.E. = 50 J = 10 n \Rightarrow n = 5$

Q 6 :

A block of mass 0.18 kg is attached to a spring of force constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = $\frac{N}{10}$ . Then N is [2011]

[IMAGE 121]

(4)

Here, loss in K.E. of the block = gain in P.E. of the spring + work done against friction

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2} + μ m g x \Rightarrow v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$

$v = \sqrt{\frac{2 \times 0.1 \times 0.18 \times 10 \times 0.6 + 2 \times 0.6 \times 0.6}{0.18}}$

$∴$ $v = \frac{4}{10} = \frac{N}{10}$ $∴$ $N = 4$

Q 7 :

A student skates up a ramp that makes an angle $30 °$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_{0}$ and wants to turn around over a semicircular path $x y z$ of radius $R$ during which he/she reaches a maximum height $h$ (at point $y$ ) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( $g$ is the acceleration due to gravity) [2020]

[IMAGE 122]

$v_{0}^{2} - 2 g h = \frac{1}{2} g R$
$v_{0}^{2} - 2 g h = \frac{\sqrt{3}}{2} g R$
the centripetal force required at points $x$ and $z$ is zero
the centripetal force required is maximum at points $x$ and $z$

1

2

3

4

Select one or more options

(1, 4)

At point Y the centripetal force provided by the component of weight mg

$∴$ $m g \sin 30 ° = \frac{m v^{2}}{R}$

$∴$ $v^{2} = \frac{g R}{2}$

Now by the energy conservation between bottom point and point Y

[IMAGE 123]

$\frac{1}{2} m v_{0}^{2} = m g h + \frac{1}{2} m v^{2}$

$∴$ $v^{2} = v_{0}^{2} - 2 g h ...(i)$

$∴$ $From eq. (i)$

$\frac{g R}{2} = v_{0}^{2} - 2 g h$

Hence option (1) is correct.

At point x and z of circular path, the points are at same height but less than h. So the velocity is more than at point y.

So required centripetal force $= \frac{m v^{2}}{r}$ is maximum at points x and y.

Q 8 :

A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $x$ -axis. Its kinetic energy $K$ changes with time as $\frac{d K}{d t} = γ t$ , where $γ$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true? [2018]

The force applied on the particle is constant
The speed of the particle is proportional to time
The distance of the particle from the origin increases linearly with time
The force is conservative

1

2

3

4

Select one or more options

(1, 2, 4)

$G \frac{d k}{d t} = γ t and k = \frac{1}{2} m V^{2}$ $∴$ $\frac{d}{d t} (\frac{1}{2} m V^{2}) = γ t$

$\Rightarrow \frac{m}{2} \times 2 V \frac{d V}{d t} = γ t$ $∴$ $m V \frac{d V}{d t} = γ t$

$∴$ $m \int_{0}^{V} V d V = γ \int_{0}^{t} t d t$ $\Rightarrow \frac{m V^{2}}{2} = \frac{γ t^{2}}{2}$

$∴$ $V = \sqrt{\frac{γ}{m}} \times t$ $i.e., V \propto t;$ $V = \frac{d s}{d t} = \sqrt{\frac{γ}{m}} t \Rightarrow s = \sqrt{\frac{γ}{m}} \cdot \frac{t^{2}}{2}$

So $V$ is proportional to $' t'$ and distance cannot be proportional to $' t'$ .

$Now F = m a = m \frac{d V}{d t} = m \frac{d}{d t} (\sqrt{\frac{γ}{m}} \times t) = m \sqrt{\frac{γ}{m}} = \sqrt{γ m} = constant$

Since force applied is constant and displacement between any two points on x-axis will also be constant, thus work done will be independent of path. Hence force is conservative in nature.

Q 9 :

A small ball starts moving from $A$ over a fixed track as shown in the figure. Surface $A B$ has friction. From $A$ to $B$ the ball rolls without slipping. Surface $B C$ is frictionless. $K_{A}$ , $K_{B}$ and $K_{C}$ are kinetic energies of the ball at A, B and C, respectively. Then [2006]

[IMAGE 124]

$h_{A} > h_{C}; K_{B} > K_{C}$
$h_{A} > h_{C}; K_{C} > K_{A}$
$h_{A} = h_{C}; K_{B} = K_{C}$
$h_{A} < h_{C}; K_{B} > K_{C}$

1

2

3

4

Select one or more options

(1, 2, 4)

From figure given in question,

Potential energy of the ball at point A $= m g h_{A}$

Potential energy of the ball at point B = 0

Potential energy of the ball at point C $= m g h_{C}$

Total energy at point A, $E_{A} = K_{A} + m g h_{A}$

Total energy at point B, $E_{B} = K_{B}$

Total energy at point C, $E_{C} = K_{C} + m g h_{C}$

As body rolls between A and B and between B and C there is no friction. So energy should be conserved here.

By law of conservation of energy

$E_{A} = E_{B} = E_{C}$

$As E_{A} = E_{C}$

$K_{A} + m g h_{A} = K_{C} + m g h_{C}$

$So, if h_{A} > h_{C} \Rightarrow K_{A} < K_{C} . So option (2) is correct.$

$If h_{A} < h_{C} \Rightarrow K_{A} > K_{C}$

$Doesn't matter if h_{A} > h_{C} or h_{A} < h_{C}, we will always have K_{B} > K_{C}$

$because E_{A} = E_{B} = E_{C} . So option (1) and (4) is also correct.$

Q 10 :

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J.

(Take the acceleration due to gravity, $g = 10 {ms}^{- 2}$ ) [2013]

[IMAGE 125]

Q. The magnitude of the normal reaction that acts on the block at the point Q is

7.5 N
8.6 N
11.5 N
22.5 N

1

2

3

4

(1)

[IMAGE 126]

$As W_{all forces} = Δ K \Rightarrow W_{m g} + W_{f r} = \frac{1}{2} m v^{2} - 0$

$\Rightarrow m g h - 150 = \frac{1}{2} m v^{2}$

$\Rightarrow m g R \sin 30 ° - 150 = \frac{1}{2} m v^{2} \Rightarrow 1 \times 10 \times 40 \times \frac{1}{2} - 150 = \frac{v^{2}}{2}$

$\Rightarrow v = 10 m/s$

$N - m g \cos θ will provide required centripetal force$

$N - m g \cos θ = \frac{m v^{2}}{R}$

$\Rightarrow N = m g \cos θ + \frac{m v^{2}}{R}$

$= 1 \times 10 \times \frac{1}{2} + \frac{1 \times {(10)}^{2}}{40} = 7.5 N$

Q 11 :

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J.

(Take the acceleration due to gravity, $g = 10 {ms}^{- 2}$ ) [2013]

[IMAGE 127]

Q. The speed of the block when it reaches the point Q is

$5 {ms}^{- 1}$
$10 {ms}^{- 1}$
$10 \sqrt{3} {ms}^{- 1}$
$20 {ms}^{- 1}$

1

2

3

4

(2)

As discussed earlier, we get $v = 10 m/s.$

Q 12 :

STATEMENT–1: A block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$ . It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30 °$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$ . The decrease in the mechanical energy in the second situation is smaller than that in the first situation.

STATEMENT–2: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. [2007]

Statement–1 is True, Statement–2 is True; Statement–2 is a correct explanation for Statement–1
Statement–1 is True, Statement–2 is True; Statement–2 is NOT a correct explanation for Statement–1
Statement–1 is True, Statement–2 is False
Statement–1 is False, Statement–2 is True

1

2

3

4

(3)

In the first case the mechanical energy is completely converted into heat because of friction, i.e., decrease in mechanical energy $= \frac{1}{2} m v^{2}$

While in second case, a part of mechanical energy is converted into heat due to friction but another part of mechanical energy is retained in the form of potential energy of the block, i.e., Decrease in mechanical energy $= \frac{1}{2} m v^{2} - m g h$

Therefore statement 1 is correct.

Statement 2 is wrong. The coefficient of friction between the block and the surface does not depend on the angle of inclination.

Topic Question Set

Q 2 :

A particle is acted by a force $F = k x$ , where $k$ is a +ve constant. Its potential energy at $x = 0$ is zero. Which curve correctly represents the variation of potential energy of the block with respect to $x$ [2004]

Q 3 :

An ideal spring with spring-constant $k$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [2002]

(4)

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 2 : A particle is acted by a force F=kx, where k is a +ve constant. Its potential energy at x=0 is zero. Which curve correctly represents the variation of potential energy of the block with respect to x [2004]

Q 3 : An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [2002]

(4)

Q 8 : A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dKdt=γt, where γ is a positive constant of appropriate dimensions. Which of the following statements is (are) true? [2018]

Q 9 : A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively. Then [2006] [IMAGE 124]

Want Us to Email you About Special Offers & Updates?

Q 2 :

A particle is acted by a force $F = k x$ , where $k$ is a +ve constant. Its potential energy at $x = 0$ is zero. Which curve correctly represents the variation of potential energy of the block with respect to $x$ [2004]

Q 3 :

An ideal spring with spring-constant $k$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [2002]