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PHXI06:WORK ENERGY AND POWER

355685 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provides equal retarding forces. Both come to rest in equal distance.
Statement B :
A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

1 Both Statement A and Statement B are correct.

2 Statement A is incorrect but Statement B is correct.

3 Both Statement A and Statement B are incorrect.

4 Statement A is correct but Statement B is incorrect.

Explanation:

$K \cdot E = \frac{1}{2} m V_{m}^{2} = \frac{1}{2} M V_{M}^{2}$
Where $m =$ Velocity of car
$M =$ Velocity of truck.
$V_{m} =$ Velocity of car
$V_{M} =$ Velocity of truck.
$∴ V_{m}^{2} = \frac{M}{m} V_{M}^{2}$
Now for the car, $∴ V_{m}^{2} = 2 \cdot a_{m} \cdot S_{m}$
$\Rightarrow S_{m} = \frac{V_{m}^{2}}{2 a_{m}} = \frac{\frac{M}{m} V_{m}^{2}}{2 \frac{F}{m}} = \frac{M V_{m}^{2}}{2 F}$ .
For the truck,
$S_{M} = \frac{V_{M}^{2}}{2 a_{M}} = \frac{V_{M}^{2}}{2 \frac{F}{M}} = \frac{M V_{M}^{2}}{2 F} = \frac{m V_{m}^{2}}{2 F}$
$∴ S_{m} = S_{M}$ .
Initial velocity of the car ${\vec{V}}_{i} = V \hat{i}$ and final velocity ${\vec{V}}_{f} = V \hat{j}$ . As velocity changes, acceleration of the car cannot be zero. So statement (4) is correct.

JEE - 2023

PHXI06:WORK ENERGY AND POWER

355686 A block of mass $0.24 k g$ is attached to a spring of force constant $4 N / m .$ The coefficient of friction between the block and the floor is $0.2 .$ 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 $0.18 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v = {(\frac{N}{100})}^{\frac{1}{2}}$ then, $N$ is (Take $g = 10 m / s^{2}$ )
supporting img

1 148

2 126

3 275

4 526

Explanation:

Decrease in mechanical energy $=$ Work done against friction
$∴ \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2} = μ m g x$
$∴ v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$
$∴ v = \sqrt{\frac{2 \times 0.2 \times 0.24 \times 10 \times (0.18) + 4 {(0.18)}^{2}}{0.24}}$
$∴ v = \sqrt{\frac{63}{50}}$
$∴ v = \sqrt{\frac{126}{100}} m / s \Rightarrow N = 126$

PHXI06:WORK ENERGY AND POWER

355687 For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1} = 4 k g$ and the horizontal surface is $μ = 0.5$ . The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_{2}$ is released from rest, when spring is in its natural length. Find minimum value of $m_{2}$ such that the block of mass $m_{1}$ will start moving.
supporting img

1

1 k g

2

3 k g

3

7 k g

4

4 k g

Explanation:

Let for just shifting of the block of mass $m_{1}$ , the elongation in spring is $x_{1}$
$∴ k x_{1} > μ m_{1} g$
$∵ x_{1 min} = \frac{μ m_{1} g}{k}$
For minimum value of $m_{2}$ , the speed of block of mass $m_{2}$ is zero when the elongation in spring is $x_{1 min}$
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_{2}$ is equal to gain in elastic potential energy in the spring.
$\begin{array}{ll} ∴ & m_{2} g x_{1 min} = \frac{1}{2} k x_{min}^{2} \\ ∴ & x_{1 min} = \frac{2 m_{2} g}{k} \\ Or & \frac{μ m_{1} g}{k} = \frac{2 m_{2} g}{k} \\ m_{2} = \frac{μ m_{1}}{2} = \frac{0.5 \times 4}{2} = 1 k g \end{array}$

PHXI06:WORK ENERGY AND POWER

355688 When a bullet of mass $10 g$ and speed $100 {ms}^{- 1}$ penetrates up to distance $1 cm$ in a human body in rest. The resistance offered by human body is:

1

2000 N

2

1500 N

3

5000 N

4

10000 N

Explanation:

Mass of bullet $= 10 g (m)$
Speed of bullet $= 100 m s^{- 1} (v)$
Distance penetrated
$= 10 c m (x) (1 c m) = 10^{- 2} m$
$U = R x \to$ displacement
$↓ ↓$
energy resistance Force
$\frac{1}{2} m v^{2} = R x$
$R = \frac{1}{2} \times \frac{10^{- 2} \times {(10^{2})}^{2}}{0.01} = 5000 N$

PHXI06:WORK ENERGY AND POWER

355689 The block of mass $m$ is pulling, vertically up with constant speed, by applying force $P$ . The free end of the string is pulled by $l$ meter, the increase in potential energy of the block is
supporting img

1

m g l

2

\frac{m g l}{2}

3

\frac{m g l}{4}

4

2 m g l

Explanation:

The constraint relation between point $P$ and block is
supporting img
The total length of the string is
$l = y_{P} + 2 y_{B} + C$
where $C$ and $l$ are constant
By differentiating the equation
$0 = Δ y_{P} + 2 Δ y_{B}$
As $(Δ y_{P}) = l \Rightarrow (Δ y_{B}) = \frac{l}{2}$
$Δ U = \frac{m g l}{2}$

PHXI06:WORK ENERGY AND POWER

355685 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provides equal retarding forces. Both come to rest in equal distance.
Statement B :
A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

1 Both Statement A and Statement B are correct.

2 Statement A is incorrect but Statement B is correct.

3 Both Statement A and Statement B are incorrect.

4 Statement A is correct but Statement B is incorrect.

Explanation:

$K \cdot E = \frac{1}{2} m V_{m}^{2} = \frac{1}{2} M V_{M}^{2}$
Where $m =$ Velocity of car
$M =$ Velocity of truck.
$V_{m} =$ Velocity of car
$V_{M} =$ Velocity of truck.
$∴ V_{m}^{2} = \frac{M}{m} V_{M}^{2}$
Now for the car, $∴ V_{m}^{2} = 2 \cdot a_{m} \cdot S_{m}$
$\Rightarrow S_{m} = \frac{V_{m}^{2}}{2 a_{m}} = \frac{\frac{M}{m} V_{m}^{2}}{2 \frac{F}{m}} = \frac{M V_{m}^{2}}{2 F}$ .
For the truck,
$S_{M} = \frac{V_{M}^{2}}{2 a_{M}} = \frac{V_{M}^{2}}{2 \frac{F}{M}} = \frac{M V_{M}^{2}}{2 F} = \frac{m V_{m}^{2}}{2 F}$
$∴ S_{m} = S_{M}$ .
Initial velocity of the car ${\vec{V}}_{i} = V \hat{i}$ and final velocity ${\vec{V}}_{f} = V \hat{j}$ . As velocity changes, acceleration of the car cannot be zero. So statement (4) is correct.

JEE - 2023

PHXI06:WORK ENERGY AND POWER

355686 A block of mass $0.24 k g$ is attached to a spring of force constant $4 N / m .$ The coefficient of friction between the block and the floor is $0.2 .$ 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 $0.18 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v = {(\frac{N}{100})}^{\frac{1}{2}}$ then, $N$ is (Take $g = 10 m / s^{2}$ )
supporting img

1 148

2 126

3 275

4 526

Explanation:

Decrease in mechanical energy $=$ Work done against friction
$∴ \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2} = μ m g x$
$∴ v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$
$∴ v = \sqrt{\frac{2 \times 0.2 \times 0.24 \times 10 \times (0.18) + 4 {(0.18)}^{2}}{0.24}}$
$∴ v = \sqrt{\frac{63}{50}}$
$∴ v = \sqrt{\frac{126}{100}} m / s \Rightarrow N = 126$

PHXI06:WORK ENERGY AND POWER

355687 For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1} = 4 k g$ and the horizontal surface is $μ = 0.5$ . The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_{2}$ is released from rest, when spring is in its natural length. Find minimum value of $m_{2}$ such that the block of mass $m_{1}$ will start moving.
supporting img

1

1 k g

2

3 k g

3

7 k g

4

4 k g

Explanation:

Let for just shifting of the block of mass $m_{1}$ , the elongation in spring is $x_{1}$
$∴ k x_{1} > μ m_{1} g$
$∵ x_{1 min} = \frac{μ m_{1} g}{k}$
For minimum value of $m_{2}$ , the speed of block of mass $m_{2}$ is zero when the elongation in spring is $x_{1 min}$
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_{2}$ is equal to gain in elastic potential energy in the spring.
$\begin{array}{ll} ∴ & m_{2} g x_{1 min} = \frac{1}{2} k x_{min}^{2} \\ ∴ & x_{1 min} = \frac{2 m_{2} g}{k} \\ Or & \frac{μ m_{1} g}{k} = \frac{2 m_{2} g}{k} \\ m_{2} = \frac{μ m_{1}}{2} = \frac{0.5 \times 4}{2} = 1 k g \end{array}$

PHXI06:WORK ENERGY AND POWER

355688 When a bullet of mass $10 g$ and speed $100 {ms}^{- 1}$ penetrates up to distance $1 cm$ in a human body in rest. The resistance offered by human body is:

1

2000 N

2

1500 N

3

5000 N

4

10000 N

Explanation:

Mass of bullet $= 10 g (m)$
Speed of bullet $= 100 m s^{- 1} (v)$
Distance penetrated
$= 10 c m (x) (1 c m) = 10^{- 2} m$
$U = R x \to$ displacement
$↓ ↓$
energy resistance Force
$\frac{1}{2} m v^{2} = R x$
$R = \frac{1}{2} \times \frac{10^{- 2} \times {(10^{2})}^{2}}{0.01} = 5000 N$

PHXI06:WORK ENERGY AND POWER

355689 The block of mass $m$ is pulling, vertically up with constant speed, by applying force $P$ . The free end of the string is pulled by $l$ meter, the increase in potential energy of the block is
supporting img

1

m g l

2

\frac{m g l}{2}

3

\frac{m g l}{4}

4

2 m g l

Explanation:

The constraint relation between point $P$ and block is
supporting img
The total length of the string is
$l = y_{P} + 2 y_{B} + C$
where $C$ and $l$ are constant
By differentiating the equation
$0 = Δ y_{P} + 2 Δ y_{B}$
As $(Δ y_{P}) = l \Rightarrow (Δ y_{B}) = \frac{l}{2}$
$Δ U = \frac{m g l}{2}$

PHXI06:WORK ENERGY AND POWER

355685 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provides equal retarding forces. Both come to rest in equal distance.
Statement B :
A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

1 Both Statement A and Statement B are correct.

2 Statement A is incorrect but Statement B is correct.

3 Both Statement A and Statement B are incorrect.

4 Statement A is correct but Statement B is incorrect.

Explanation:

$K \cdot E = \frac{1}{2} m V_{m}^{2} = \frac{1}{2} M V_{M}^{2}$
Where $m =$ Velocity of car
$M =$ Velocity of truck.
$V_{m} =$ Velocity of car
$V_{M} =$ Velocity of truck.
$∴ V_{m}^{2} = \frac{M}{m} V_{M}^{2}$
Now for the car, $∴ V_{m}^{2} = 2 \cdot a_{m} \cdot S_{m}$
$\Rightarrow S_{m} = \frac{V_{m}^{2}}{2 a_{m}} = \frac{\frac{M}{m} V_{m}^{2}}{2 \frac{F}{m}} = \frac{M V_{m}^{2}}{2 F}$ .
For the truck,
$S_{M} = \frac{V_{M}^{2}}{2 a_{M}} = \frac{V_{M}^{2}}{2 \frac{F}{M}} = \frac{M V_{M}^{2}}{2 F} = \frac{m V_{m}^{2}}{2 F}$
$∴ S_{m} = S_{M}$ .
Initial velocity of the car ${\vec{V}}_{i} = V \hat{i}$ and final velocity ${\vec{V}}_{f} = V \hat{j}$ . As velocity changes, acceleration of the car cannot be zero. So statement (4) is correct.

JEE - 2023

PHXI06:WORK ENERGY AND POWER

355686 A block of mass $0.24 k g$ is attached to a spring of force constant $4 N / m .$ The coefficient of friction between the block and the floor is $0.2 .$ 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 $0.18 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v = {(\frac{N}{100})}^{\frac{1}{2}}$ then, $N$ is (Take $g = 10 m / s^{2}$ )
supporting img

1 148

2 126

3 275

4 526

Explanation:

Decrease in mechanical energy $=$ Work done against friction
$∴ \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2} = μ m g x$
$∴ v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$
$∴ v = \sqrt{\frac{2 \times 0.2 \times 0.24 \times 10 \times (0.18) + 4 {(0.18)}^{2}}{0.24}}$
$∴ v = \sqrt{\frac{63}{50}}$
$∴ v = \sqrt{\frac{126}{100}} m / s \Rightarrow N = 126$

PHXI06:WORK ENERGY AND POWER

355687 For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1} = 4 k g$ and the horizontal surface is $μ = 0.5$ . The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_{2}$ is released from rest, when spring is in its natural length. Find minimum value of $m_{2}$ such that the block of mass $m_{1}$ will start moving.
supporting img

1

1 k g

2

3 k g

3

7 k g

4

4 k g

Explanation:

Let for just shifting of the block of mass $m_{1}$ , the elongation in spring is $x_{1}$
$∴ k x_{1} > μ m_{1} g$
$∵ x_{1 min} = \frac{μ m_{1} g}{k}$
For minimum value of $m_{2}$ , the speed of block of mass $m_{2}$ is zero when the elongation in spring is $x_{1 min}$
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_{2}$ is equal to gain in elastic potential energy in the spring.
$\begin{array}{ll} ∴ & m_{2} g x_{1 min} = \frac{1}{2} k x_{min}^{2} \\ ∴ & x_{1 min} = \frac{2 m_{2} g}{k} \\ Or & \frac{μ m_{1} g}{k} = \frac{2 m_{2} g}{k} \\ m_{2} = \frac{μ m_{1}}{2} = \frac{0.5 \times 4}{2} = 1 k g \end{array}$

PHXI06:WORK ENERGY AND POWER

355688 When a bullet of mass $10 g$ and speed $100 {ms}^{- 1}$ penetrates up to distance $1 cm$ in a human body in rest. The resistance offered by human body is:

1

2000 N

2

1500 N

3

5000 N

4

10000 N

Explanation:

Mass of bullet $= 10 g (m)$
Speed of bullet $= 100 m s^{- 1} (v)$
Distance penetrated
$= 10 c m (x) (1 c m) = 10^{- 2} m$
$U = R x \to$ displacement
$↓ ↓$
energy resistance Force
$\frac{1}{2} m v^{2} = R x$
$R = \frac{1}{2} \times \frac{10^{- 2} \times {(10^{2})}^{2}}{0.01} = 5000 N$

PHXI06:WORK ENERGY AND POWER

355689 The block of mass $m$ is pulling, vertically up with constant speed, by applying force $P$ . The free end of the string is pulled by $l$ meter, the increase in potential energy of the block is
supporting img

1

m g l

2

\frac{m g l}{2}

3

\frac{m g l}{4}

4

2 m g l

Explanation:

The constraint relation between point $P$ and block is
supporting img
The total length of the string is
$l = y_{P} + 2 y_{B} + C$
where $C$ and $l$ are constant
By differentiating the equation
$0 = Δ y_{P} + 2 Δ y_{B}$
As $(Δ y_{P}) = l \Rightarrow (Δ y_{B}) = \frac{l}{2}$
$Δ U = \frac{m g l}{2}$

PHXI06:WORK ENERGY AND POWER

355685 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provides equal retarding forces. Both come to rest in equal distance.
Statement B :
A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

1 Both Statement A and Statement B are correct.

2 Statement A is incorrect but Statement B is correct.

3 Both Statement A and Statement B are incorrect.

4 Statement A is correct but Statement B is incorrect.

Explanation:

$K \cdot E = \frac{1}{2} m V_{m}^{2} = \frac{1}{2} M V_{M}^{2}$
Where $m =$ Velocity of car
$M =$ Velocity of truck.
$V_{m} =$ Velocity of car
$V_{M} =$ Velocity of truck.
$∴ V_{m}^{2} = \frac{M}{m} V_{M}^{2}$
Now for the car, $∴ V_{m}^{2} = 2 \cdot a_{m} \cdot S_{m}$
$\Rightarrow S_{m} = \frac{V_{m}^{2}}{2 a_{m}} = \frac{\frac{M}{m} V_{m}^{2}}{2 \frac{F}{m}} = \frac{M V_{m}^{2}}{2 F}$ .
For the truck,
$S_{M} = \frac{V_{M}^{2}}{2 a_{M}} = \frac{V_{M}^{2}}{2 \frac{F}{M}} = \frac{M V_{M}^{2}}{2 F} = \frac{m V_{m}^{2}}{2 F}$
$∴ S_{m} = S_{M}$ .
Initial velocity of the car ${\vec{V}}_{i} = V \hat{i}$ and final velocity ${\vec{V}}_{f} = V \hat{j}$ . As velocity changes, acceleration of the car cannot be zero. So statement (4) is correct.

JEE - 2023

PHXI06:WORK ENERGY AND POWER

355686 A block of mass $0.24 k g$ is attached to a spring of force constant $4 N / m .$ The coefficient of friction between the block and the floor is $0.2 .$ 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 $0.18 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v = {(\frac{N}{100})}^{\frac{1}{2}}$ then, $N$ is (Take $g = 10 m / s^{2}$ )
supporting img

1 148

2 126

3 275

4 526

Explanation:

Decrease in mechanical energy $=$ Work done against friction
$∴ \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2} = μ m g x$
$∴ v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$
$∴ v = \sqrt{\frac{2 \times 0.2 \times 0.24 \times 10 \times (0.18) + 4 {(0.18)}^{2}}{0.24}}$
$∴ v = \sqrt{\frac{63}{50}}$
$∴ v = \sqrt{\frac{126}{100}} m / s \Rightarrow N = 126$

PHXI06:WORK ENERGY AND POWER

355687 For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1} = 4 k g$ and the horizontal surface is $μ = 0.5$ . The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_{2}$ is released from rest, when spring is in its natural length. Find minimum value of $m_{2}$ such that the block of mass $m_{1}$ will start moving.
supporting img

1

1 k g

2

3 k g

3

7 k g

4

4 k g

Explanation:

Let for just shifting of the block of mass $m_{1}$ , the elongation in spring is $x_{1}$
$∴ k x_{1} > μ m_{1} g$
$∵ x_{1 min} = \frac{μ m_{1} g}{k}$
For minimum value of $m_{2}$ , the speed of block of mass $m_{2}$ is zero when the elongation in spring is $x_{1 min}$
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_{2}$ is equal to gain in elastic potential energy in the spring.
$\begin{array}{ll} ∴ & m_{2} g x_{1 min} = \frac{1}{2} k x_{min}^{2} \\ ∴ & x_{1 min} = \frac{2 m_{2} g}{k} \\ Or & \frac{μ m_{1} g}{k} = \frac{2 m_{2} g}{k} \\ m_{2} = \frac{μ m_{1}}{2} = \frac{0.5 \times 4}{2} = 1 k g \end{array}$

PHXI06:WORK ENERGY AND POWER

355688 When a bullet of mass $10 g$ and speed $100 {ms}^{- 1}$ penetrates up to distance $1 cm$ in a human body in rest. The resistance offered by human body is:

1

2000 N

2

1500 N

3

5000 N

4

10000 N

Explanation:

Mass of bullet $= 10 g (m)$
Speed of bullet $= 100 m s^{- 1} (v)$
Distance penetrated
$= 10 c m (x) (1 c m) = 10^{- 2} m$
$U = R x \to$ displacement
$↓ ↓$
energy resistance Force
$\frac{1}{2} m v^{2} = R x$
$R = \frac{1}{2} \times \frac{10^{- 2} \times {(10^{2})}^{2}}{0.01} = 5000 N$

PHXI06:WORK ENERGY AND POWER

355689 The block of mass $m$ is pulling, vertically up with constant speed, by applying force $P$ . The free end of the string is pulled by $l$ meter, the increase in potential energy of the block is
supporting img

1

m g l

2

\frac{m g l}{2}

3

\frac{m g l}{4}

4

2 m g l

Explanation:

The constraint relation between point $P$ and block is
supporting img
The total length of the string is
$l = y_{P} + 2 y_{B} + C$
where $C$ and $l$ are constant
By differentiating the equation
$0 = Δ y_{P} + 2 Δ y_{B}$
As $(Δ y_{P}) = l \Rightarrow (Δ y_{B}) = \frac{l}{2}$
$Δ U = \frac{m g l}{2}$

PHXI06:WORK ENERGY AND POWER

355685 Read the Statement - A and Statement - B carefully to mark the correct options given below:
Statement A :
A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provides equal retarding forces. Both come to rest in equal distance.
Statement B :
A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.

1 Both Statement A and Statement B are correct.

2 Statement A is incorrect but Statement B is correct.

3 Both Statement A and Statement B are incorrect.

4 Statement A is correct but Statement B is incorrect.

Explanation:

$K \cdot E = \frac{1}{2} m V_{m}^{2} = \frac{1}{2} M V_{M}^{2}$
Where $m =$ Velocity of car
$M =$ Velocity of truck.
$V_{m} =$ Velocity of car
$V_{M} =$ Velocity of truck.
$∴ V_{m}^{2} = \frac{M}{m} V_{M}^{2}$
Now for the car, $∴ V_{m}^{2} = 2 \cdot a_{m} \cdot S_{m}$
$\Rightarrow S_{m} = \frac{V_{m}^{2}}{2 a_{m}} = \frac{\frac{M}{m} V_{m}^{2}}{2 \frac{F}{m}} = \frac{M V_{m}^{2}}{2 F}$ .
For the truck,
$S_{M} = \frac{V_{M}^{2}}{2 a_{M}} = \frac{V_{M}^{2}}{2 \frac{F}{M}} = \frac{M V_{M}^{2}}{2 F} = \frac{m V_{m}^{2}}{2 F}$
$∴ S_{m} = S_{M}$ .
Initial velocity of the car ${\vec{V}}_{i} = V \hat{i}$ and final velocity ${\vec{V}}_{f} = V \hat{j}$ . As velocity changes, acceleration of the car cannot be zero. So statement (4) is correct.

JEE - 2023

PHXI06:WORK ENERGY AND POWER

355686 A block of mass $0.24 k g$ is attached to a spring of force constant $4 N / m .$ The coefficient of friction between the block and the floor is $0.2 .$ 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 $0.18 m$ and comes to rest for the first time. The initial velocity of the block in $m / s$ is $v = {(\frac{N}{100})}^{\frac{1}{2}}$ then, $N$ is (Take $g = 10 m / s^{2}$ )
supporting img

1 148

2 126

3 275

4 526

Explanation:

Decrease in mechanical energy $=$ Work done against friction
$∴ \frac{1}{2} m v^{2} - \frac{1}{2} k x^{2} = μ m g x$
$∴ v = \sqrt{\frac{2 μ m g x + k x^{2}}{m}}$
$∴ v = \sqrt{\frac{2 \times 0.2 \times 0.24 \times 10 \times (0.18) + 4 {(0.18)}^{2}}{0.24}}$
$∴ v = \sqrt{\frac{63}{50}}$
$∴ v = \sqrt{\frac{126}{100}} m / s \Rightarrow N = 126$

PHXI06:WORK ENERGY AND POWER

355687 For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1} = 4 k g$ and the horizontal surface is $μ = 0.5$ . The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_{2}$ is released from rest, when spring is in its natural length. Find minimum value of $m_{2}$ such that the block of mass $m_{1}$ will start moving.
supporting img

1

1 k g

2

3 k g

3

7 k g

4

4 k g

Explanation:

Let for just shifting of the block of mass $m_{1}$ , the elongation in spring is $x_{1}$
$∴ k x_{1} > μ m_{1} g$
$∵ x_{1 min} = \frac{μ m_{1} g}{k}$
For minimum value of $m_{2}$ , the speed of block of mass $m_{2}$ is zero when the elongation in spring is $x_{1 min}$
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_{2}$ is equal to gain in elastic potential energy in the spring.
$\begin{array}{ll} ∴ & m_{2} g x_{1 min} = \frac{1}{2} k x_{min}^{2} \\ ∴ & x_{1 min} = \frac{2 m_{2} g}{k} \\ Or & \frac{μ m_{1} g}{k} = \frac{2 m_{2} g}{k} \\ m_{2} = \frac{μ m_{1}}{2} = \frac{0.5 \times 4}{2} = 1 k g \end{array}$

PHXI06:WORK ENERGY AND POWER

355688 When a bullet of mass $10 g$ and speed $100 {ms}^{- 1}$ penetrates up to distance $1 cm$ in a human body in rest. The resistance offered by human body is:

1

2000 N

2

1500 N

3

5000 N

4

10000 N

Explanation:

Mass of bullet $= 10 g (m)$
Speed of bullet $= 100 m s^{- 1} (v)$
Distance penetrated
$= 10 c m (x) (1 c m) = 10^{- 2} m$
$U = R x \to$ displacement
$↓ ↓$
energy resistance Force
$\frac{1}{2} m v^{2} = R x$
$R = \frac{1}{2} \times \frac{10^{- 2} \times {(10^{2})}^{2}}{0.01} = 5000 N$

PHXI06:WORK ENERGY AND POWER

355689 The block of mass $m$ is pulling, vertically up with constant speed, by applying force $P$ . The free end of the string is pulled by $l$ meter, the increase in potential energy of the block is
supporting img

1

m g l

2

\frac{m g l}{2}

3

\frac{m g l}{4}

4

2 m g l

Explanation:

The constraint relation between point $P$ and block is
supporting img
The total length of the string is
$l = y_{P} + 2 y_{B} + C$
where $C$ and $l$ are constant
By differentiating the equation
$0 = Δ y_{P} + 2 Δ y_{B}$
As $(Δ y_{P}) = l \Rightarrow (Δ y_{B}) = \frac{l}{2}$
$Δ U = \frac{m g l}{2}$