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PHXI06:WORK ENERGY AND POWER

355342 The total mechanical energy of sun-planet system is:

1

+ v e

2

- v e

3 Zero

4 Equal to its potential energy

PHXI06:WORK ENERGY AND POWER

355343 Initially spring in its natural length, the block of mass $0.25 k g$ is released, then find out the value of maximum force by system on the floor
supporting img

1

15 N

2

20 N

3

25 N

4

30 N

Explanation:

If $x$ be the compression in the spring, when a block of $0.25 k g$ is released.
From energy conservation,
Potential energy of spring $=$ Potential energy of block
$\Rightarrow \frac{1}{2} k x^{2} = m g x$
Where, $k$ is force constant of spring.
$k x = 2 m g = 2 \times 0.25 g$
$k x = 0.5 g$
i.e. Force applied by the spring on the block,
$F = k x$
$\Rightarrow F = 0.5 g$
From Free body diagram,
supporting img
$∴$ Maximum force by system on the floor,
$N = F + 2 g$
$\begin{aligned} = 0.5 g + 2 g = 2.5 g (∵ g = 10 m^{- 2}) \\ = 25 N \end{aligned}$

AIIMS - 2019

PHXI06:WORK ENERGY AND POWER

355344 A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
supporting img

1

1 m

2

\sqrt{6} m

3

2 m

4

\sqrt{5} m

Explanation:

supporting img

$h = 10 \sin 30 = 5 m$
Let the maximum compression of spring be $x$ .
Using the work-energy theorem,
$Δ K =$ Work done by all forces
$Δ K = m g h - f_{K} \times 2 - \frac{1}{2} K x^{2} = 0$
$[∵ K E_{i} = K E_{f} = 0]$
Here $f_{K} = μ_{K} N = μ_{K} m g = 0.5 \times 5 \times 10 = 25 N$
$5 \times 10 \times 5 - 25 \times 2 - \frac{1}{2} \times 100 x^{2} = 0$
On solving, we get $x = 2 m$

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355345 A machine which is $75 %$ efficient, uses 12 $J$ of energy in lifting 1 $k g$ mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

1

\sqrt{32} m s^{- 1}

2

\sqrt{24} m s^{- 1}

3

\sqrt{18} m s^{- 1}

4

\sqrt{12} m s^{- 1}

Explanation:

From energy conservation $0.75 (12 J) = m g h = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{18} m / s$

PHXI06:WORK ENERGY AND POWER

355342 The total mechanical energy of sun-planet system is:

1

+ v e

2

- v e

3 Zero

4 Equal to its potential energy

PHXI06:WORK ENERGY AND POWER

355343 Initially spring in its natural length, the block of mass $0.25 k g$ is released, then find out the value of maximum force by system on the floor
supporting img

1

15 N

2

20 N

3

25 N

4

30 N

Explanation:

If $x$ be the compression in the spring, when a block of $0.25 k g$ is released.
From energy conservation,
Potential energy of spring $=$ Potential energy of block
$\Rightarrow \frac{1}{2} k x^{2} = m g x$
Where, $k$ is force constant of spring.
$k x = 2 m g = 2 \times 0.25 g$
$k x = 0.5 g$
i.e. Force applied by the spring on the block,
$F = k x$
$\Rightarrow F = 0.5 g$
From Free body diagram,
supporting img
$∴$ Maximum force by system on the floor,
$N = F + 2 g$
$\begin{aligned} = 0.5 g + 2 g = 2.5 g (∵ g = 10 m^{- 2}) \\ = 25 N \end{aligned}$

AIIMS - 2019

PHXI06:WORK ENERGY AND POWER

355344 A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
supporting img

1

1 m

2

\sqrt{6} m

3

2 m

4

\sqrt{5} m

Explanation:

supporting img

$h = 10 \sin 30 = 5 m$
Let the maximum compression of spring be $x$ .
Using the work-energy theorem,
$Δ K =$ Work done by all forces
$Δ K = m g h - f_{K} \times 2 - \frac{1}{2} K x^{2} = 0$
$[∵ K E_{i} = K E_{f} = 0]$
Here $f_{K} = μ_{K} N = μ_{K} m g = 0.5 \times 5 \times 10 = 25 N$
$5 \times 10 \times 5 - 25 \times 2 - \frac{1}{2} \times 100 x^{2} = 0$
On solving, we get $x = 2 m$

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355345 A machine which is $75 %$ efficient, uses 12 $J$ of energy in lifting 1 $k g$ mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

1

\sqrt{32} m s^{- 1}

2

\sqrt{24} m s^{- 1}

3

\sqrt{18} m s^{- 1}

4

\sqrt{12} m s^{- 1}

Explanation:

From energy conservation $0.75 (12 J) = m g h = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{18} m / s$

PHXI06:WORK ENERGY AND POWER

355342 The total mechanical energy of sun-planet system is:

1

+ v e

2

- v e

3 Zero

4 Equal to its potential energy

PHXI06:WORK ENERGY AND POWER

355343 Initially spring in its natural length, the block of mass $0.25 k g$ is released, then find out the value of maximum force by system on the floor
supporting img

1

15 N

2

20 N

3

25 N

4

30 N

Explanation:

If $x$ be the compression in the spring, when a block of $0.25 k g$ is released.
From energy conservation,
Potential energy of spring $=$ Potential energy of block
$\Rightarrow \frac{1}{2} k x^{2} = m g x$
Where, $k$ is force constant of spring.
$k x = 2 m g = 2 \times 0.25 g$
$k x = 0.5 g$
i.e. Force applied by the spring on the block,
$F = k x$
$\Rightarrow F = 0.5 g$
From Free body diagram,
supporting img
$∴$ Maximum force by system on the floor,
$N = F + 2 g$
$\begin{aligned} = 0.5 g + 2 g = 2.5 g (∵ g = 10 m^{- 2}) \\ = 25 N \end{aligned}$

AIIMS - 2019

PHXI06:WORK ENERGY AND POWER

355344 A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
supporting img

1

1 m

2

\sqrt{6} m

3

2 m

4

\sqrt{5} m

Explanation:

supporting img

$h = 10 \sin 30 = 5 m$
Let the maximum compression of spring be $x$ .
Using the work-energy theorem,
$Δ K =$ Work done by all forces
$Δ K = m g h - f_{K} \times 2 - \frac{1}{2} K x^{2} = 0$
$[∵ K E_{i} = K E_{f} = 0]$
Here $f_{K} = μ_{K} N = μ_{K} m g = 0.5 \times 5 \times 10 = 25 N$
$5 \times 10 \times 5 - 25 \times 2 - \frac{1}{2} \times 100 x^{2} = 0$
On solving, we get $x = 2 m$

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355345 A machine which is $75 %$ efficient, uses 12 $J$ of energy in lifting 1 $k g$ mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

1

\sqrt{32} m s^{- 1}

2

\sqrt{24} m s^{- 1}

3

\sqrt{18} m s^{- 1}

4

\sqrt{12} m s^{- 1}

Explanation:

From energy conservation $0.75 (12 J) = m g h = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{18} m / s$

PHXI06:WORK ENERGY AND POWER

355342 The total mechanical energy of sun-planet system is:

1

+ v e

2

- v e

3 Zero

4 Equal to its potential energy

PHXI06:WORK ENERGY AND POWER

355343 Initially spring in its natural length, the block of mass $0.25 k g$ is released, then find out the value of maximum force by system on the floor
supporting img

1

15 N

2

20 N

3

25 N

4

30 N

Explanation:

If $x$ be the compression in the spring, when a block of $0.25 k g$ is released.
From energy conservation,
Potential energy of spring $=$ Potential energy of block
$\Rightarrow \frac{1}{2} k x^{2} = m g x$
Where, $k$ is force constant of spring.
$k x = 2 m g = 2 \times 0.25 g$
$k x = 0.5 g$
i.e. Force applied by the spring on the block,
$F = k x$
$\Rightarrow F = 0.5 g$
From Free body diagram,
supporting img
$∴$ Maximum force by system on the floor,
$N = F + 2 g$
$\begin{aligned} = 0.5 g + 2 g = 2.5 g (∵ g = 10 m^{- 2}) \\ = 25 N \end{aligned}$

AIIMS - 2019

PHXI06:WORK ENERGY AND POWER

355344 A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is
supporting img

1

1 m

2

\sqrt{6} m

3

2 m

4

\sqrt{5} m

Explanation:

supporting img

$h = 10 \sin 30 = 5 m$
Let the maximum compression of spring be $x$ .
Using the work-energy theorem,
$Δ K =$ Work done by all forces
$Δ K = m g h - f_{K} \times 2 - \frac{1}{2} K x^{2} = 0$
$[∵ K E_{i} = K E_{f} = 0]$
Here $f_{K} = μ_{K} N = μ_{K} m g = 0.5 \times 5 \times 10 = 25 N$
$5 \times 10 \times 5 - 25 \times 2 - \frac{1}{2} \times 100 x^{2} = 0$
On solving, we get $x = 2 m$

JEE - 2024

PHXI06:WORK ENERGY AND POWER

355345 A machine which is $75 %$ efficient, uses 12 $J$ of energy in lifting 1 $k g$ mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

1

\sqrt{32} m s^{- 1}

2

\sqrt{24} m s^{- 1}

3

\sqrt{18} m s^{- 1}

4

\sqrt{12} m s^{- 1}

Explanation:

From energy conservation $0.75 (12 J) = m g h = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{18} m / s$

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