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Work, Energy and Power

268873 A particle of mass $100 g$ is thrown vertically upwards with a speed of $5 m / s$ . The work done by the force of gravity during the time the particle goes up is

1

- 0.5 J

2

- 1.25 J

3

1.25 J

4

0.5 J

Explanation:

$h = \frac{u^{2}}{2 g}$ ; workdone by gravity $= F S \cos θ$
$here F = m g, S = h, θ = 180^{\circ}$

Work, Energy and Power

268874 A large slab of mass $5 kg$ lies on a smooth horizontal surface, with a block of mass $4 kg$ lying on the top of it. The coefficient of friction between the block and the slab is 0.25 . If the block is pulled horizontally by a force of $F = 6 N$ , then the work done by the force of friction on the slab, between the instants $t = 2 s$ to $t = 3 s$ is $(g = 10 {ms}^{- 2})$

1

2.4 J

2

5.55 J

3

4 .44 J

4

10 J

Explanation:

Maximum frictional force between the slab and the block
$f_{max} = μ N = μ mg; m =$ mass of upper block
Evidently, \(Fso, the two bodies will move together as a single unit. If ' $a$ ' be their combined acceleration, then
$a = \frac{F}{m + M}$
Therefore, frictional force acting can be obtained as $f = M a$ , Using $s = \frac{1}{2} a t^{2}$ , find $S_{2}$ and $S_{3}$ . $W = (S_{3} - S_{2}) f$

Work, Energy and Power

268875 In the pulley - block system shown in figure, strings are light. Pulleys are massless and smooth. System is released from rest. In 0.3 seconds

a) work done on $2 kg$ block by gravity is $6 J$
b) work done on $2 kg$ block by string is $- 2 J$
c) work done on $1 kg$ block by gravity is $- 1.5 J$
d) work done on $1 kg$ block by string is $2 J$

1 only a, d are correct

2 only b, d are correct

3 only a, b, c are correct

4 All are correct

Explanation:

On $2 kg$ block , $W_{1} = m_{1} g x = m_{1} g ◻ ◻^{1} a t^{2} ⊟$ ;
$W_{T} = T \times x = T ◻ \frac{1}{2} a t^{2}$ ,
On $1 kg$ block, $W_{2} = m_{2} g \frac{x}{2}$

Work, Energy and Power

268876 A body of mass $0.5 kg$ travels in a straight line with a velocity $_{v = a x^{\frac{3}{2}}}$ where $a = 5 m^{- \frac{1}{2}} s^{- 1}$ . What is the work done by the net force during its displacement from $x = 0$ to $x = 2 m$ ?P.E AND K.E

1

50 J

2

20 J

3

80 J

4

45.5 J

Explanation:

$W = Δ K E = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

Work, Energy and Power

268873 A particle of mass $100 g$ is thrown vertically upwards with a speed of $5 m / s$ . The work done by the force of gravity during the time the particle goes up is

1

- 0.5 J

2

- 1.25 J

3

1.25 J

4

0.5 J

Explanation:

$h = \frac{u^{2}}{2 g}$ ; workdone by gravity $= F S \cos θ$
$here F = m g, S = h, θ = 180^{\circ}$

Work, Energy and Power

268874 A large slab of mass $5 kg$ lies on a smooth horizontal surface, with a block of mass $4 kg$ lying on the top of it. The coefficient of friction between the block and the slab is 0.25 . If the block is pulled horizontally by a force of $F = 6 N$ , then the work done by the force of friction on the slab, between the instants $t = 2 s$ to $t = 3 s$ is $(g = 10 {ms}^{- 2})$

1

2.4 J

2

5.55 J

3

4 .44 J

4

10 J

Explanation:

Maximum frictional force between the slab and the block
$f_{max} = μ N = μ mg; m =$ mass of upper block
Evidently, \(Fso, the two bodies will move together as a single unit. If ' $a$ ' be their combined acceleration, then
$a = \frac{F}{m + M}$
Therefore, frictional force acting can be obtained as $f = M a$ , Using $s = \frac{1}{2} a t^{2}$ , find $S_{2}$ and $S_{3}$ . $W = (S_{3} - S_{2}) f$

Work, Energy and Power

268875 In the pulley - block system shown in figure, strings are light. Pulleys are massless and smooth. System is released from rest. In 0.3 seconds

a) work done on $2 kg$ block by gravity is $6 J$
b) work done on $2 kg$ block by string is $- 2 J$
c) work done on $1 kg$ block by gravity is $- 1.5 J$
d) work done on $1 kg$ block by string is $2 J$

1 only a, d are correct

2 only b, d are correct

3 only a, b, c are correct

4 All are correct

Explanation:

On $2 kg$ block , $W_{1} = m_{1} g x = m_{1} g ◻ ◻^{1} a t^{2} ⊟$ ;
$W_{T} = T \times x = T ◻ \frac{1}{2} a t^{2}$ ,
On $1 kg$ block, $W_{2} = m_{2} g \frac{x}{2}$

Work, Energy and Power

268876 A body of mass $0.5 kg$ travels in a straight line with a velocity $_{v = a x^{\frac{3}{2}}}$ where $a = 5 m^{- \frac{1}{2}} s^{- 1}$ . What is the work done by the net force during its displacement from $x = 0$ to $x = 2 m$ ?P.E AND K.E

1

50 J

2

20 J

3

80 J

4

45.5 J

Explanation:

$W = Δ K E = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

Work, Energy and Power

268873 A particle of mass $100 g$ is thrown vertically upwards with a speed of $5 m / s$ . The work done by the force of gravity during the time the particle goes up is

1

- 0.5 J

2

- 1.25 J

3

1.25 J

4

0.5 J

Explanation:

$h = \frac{u^{2}}{2 g}$ ; workdone by gravity $= F S \cos θ$
$here F = m g, S = h, θ = 180^{\circ}$

Work, Energy and Power

268874 A large slab of mass $5 kg$ lies on a smooth horizontal surface, with a block of mass $4 kg$ lying on the top of it. The coefficient of friction between the block and the slab is 0.25 . If the block is pulled horizontally by a force of $F = 6 N$ , then the work done by the force of friction on the slab, between the instants $t = 2 s$ to $t = 3 s$ is $(g = 10 {ms}^{- 2})$

1

2.4 J

2

5.55 J

3

4 .44 J

4

10 J

Explanation:

Maximum frictional force between the slab and the block
$f_{max} = μ N = μ mg; m =$ mass of upper block
Evidently, \(Fso, the two bodies will move together as a single unit. If ' $a$ ' be their combined acceleration, then
$a = \frac{F}{m + M}$
Therefore, frictional force acting can be obtained as $f = M a$ , Using $s = \frac{1}{2} a t^{2}$ , find $S_{2}$ and $S_{3}$ . $W = (S_{3} - S_{2}) f$

Work, Energy and Power

268875 In the pulley - block system shown in figure, strings are light. Pulleys are massless and smooth. System is released from rest. In 0.3 seconds

a) work done on $2 kg$ block by gravity is $6 J$
b) work done on $2 kg$ block by string is $- 2 J$
c) work done on $1 kg$ block by gravity is $- 1.5 J$
d) work done on $1 kg$ block by string is $2 J$

1 only a, d are correct

2 only b, d are correct

3 only a, b, c are correct

4 All are correct

Explanation:

On $2 kg$ block , $W_{1} = m_{1} g x = m_{1} g ◻ ◻^{1} a t^{2} ⊟$ ;
$W_{T} = T \times x = T ◻ \frac{1}{2} a t^{2}$ ,
On $1 kg$ block, $W_{2} = m_{2} g \frac{x}{2}$

Work, Energy and Power

268876 A body of mass $0.5 kg$ travels in a straight line with a velocity $_{v = a x^{\frac{3}{2}}}$ where $a = 5 m^{- \frac{1}{2}} s^{- 1}$ . What is the work done by the net force during its displacement from $x = 0$ to $x = 2 m$ ?P.E AND K.E

1

50 J

2

20 J

3

80 J

4

45.5 J

Explanation:

$W = Δ K E = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

Work, Energy and Power

268873 A particle of mass $100 g$ is thrown vertically upwards with a speed of $5 m / s$ . The work done by the force of gravity during the time the particle goes up is

1

- 0.5 J

2

- 1.25 J

3

1.25 J

4

0.5 J

Explanation:

$h = \frac{u^{2}}{2 g}$ ; workdone by gravity $= F S \cos θ$
$here F = m g, S = h, θ = 180^{\circ}$

Work, Energy and Power

268874 A large slab of mass $5 kg$ lies on a smooth horizontal surface, with a block of mass $4 kg$ lying on the top of it. The coefficient of friction between the block and the slab is 0.25 . If the block is pulled horizontally by a force of $F = 6 N$ , then the work done by the force of friction on the slab, between the instants $t = 2 s$ to $t = 3 s$ is $(g = 10 {ms}^{- 2})$

1

2.4 J

2

5.55 J

3

4 .44 J

4

10 J

Explanation:

Maximum frictional force between the slab and the block
$f_{max} = μ N = μ mg; m =$ mass of upper block
Evidently, \(Fso, the two bodies will move together as a single unit. If ' $a$ ' be their combined acceleration, then
$a = \frac{F}{m + M}$
Therefore, frictional force acting can be obtained as $f = M a$ , Using $s = \frac{1}{2} a t^{2}$ , find $S_{2}$ and $S_{3}$ . $W = (S_{3} - S_{2}) f$

Work, Energy and Power

268875 In the pulley - block system shown in figure, strings are light. Pulleys are massless and smooth. System is released from rest. In 0.3 seconds

a) work done on $2 kg$ block by gravity is $6 J$
b) work done on $2 kg$ block by string is $- 2 J$
c) work done on $1 kg$ block by gravity is $- 1.5 J$
d) work done on $1 kg$ block by string is $2 J$

1 only a, d are correct

2 only b, d are correct

3 only a, b, c are correct

4 All are correct

Explanation:

On $2 kg$ block , $W_{1} = m_{1} g x = m_{1} g ◻ ◻^{1} a t^{2} ⊟$ ;
$W_{T} = T \times x = T ◻ \frac{1}{2} a t^{2}$ ,
On $1 kg$ block, $W_{2} = m_{2} g \frac{x}{2}$

Work, Energy and Power

268876 A body of mass $0.5 kg$ travels in a straight line with a velocity $_{v = a x^{\frac{3}{2}}}$ where $a = 5 m^{- \frac{1}{2}} s^{- 1}$ . What is the work done by the net force during its displacement from $x = 0$ to $x = 2 m$ ?P.E AND K.E

1

50 J

2

20 J

3

80 J

4

45.5 J

Explanation:

$W = Δ K E = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

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