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Work, Energy and Power

268705 A tennis ball has a mass of $56.7 gm$ and is served by a player with a speed of $180 kmph$ . The work done in serving the ball is nearly

1

710 J

2

71 J

3

918 J

4

91.8 J

Explanation:

$W = \frac{1}{2} m v_{i}^{2};$

Work, Energy and Power

268706 A body of mass $2 kg$ is projected vertically up with velocity $5 {ms}^{- 1}$ . The work done on the body by gravitational force before it is brought to rest momentarily is

1

250 J

2

25 J

3

0 J

4

- 25 J

Explanation:

$W = - m g h = - \frac{1}{2} m u^{2}$

Work, Energy and Power

268750 A body of mass $5 k g$ is moved up over $10 m$ along the line of greatest slope of a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. If $g = 10 m / s^{2}$ , the work done will be

1

500 J

2

2500 J

3

250 J

4

25 J

Explanation:

$W = \vec{F} \cdot \vec{S} = FS \cos θ$ ; here $F = mgsin α$ and
$θ = 0^{\circ}, α = 30^{\circ}$

Work, Energy and Power

268751 A particle of mass $0.5 kg$ is displaced from position $\vec{r_{1}} (2, 3, 1)$ to $\vec{r_{2}} (4, 3, 2)$ by applying a force of magnitude $30 N$ which is acting along $\exists i + \hat{j} + \hat{k} ⊟$ . The work done by the force is

1

10 \sqrt{3} J

2

30 \sqrt{3} J

3

30 J

4

40 J

Explanation:

$F_{x} = F \cos α, F_{y} = F \cos β, F_{z} = F \cos γ$ ,
$W = F . S$

Work, Energy and Power

268752 Kinetic energy of a particle moving in a straight line varies with time $t$ as $K = 4 t^{2}$ . The force acting on the particle

1 is constant

2 is increasing

3 is decreasing

4 first increases and then decreases

Explanation:

$\frac{1}{2} m v^{2} = 4 t^{2}; ∴ v = \sqrt{\frac{8}{m}} t$
comparing with $v = at, a =$ constant
i.e., the force acting on the particle is constant

Work, Energy and Power

268705 A tennis ball has a mass of $56.7 gm$ and is served by a player with a speed of $180 kmph$ . The work done in serving the ball is nearly

1

710 J

2

71 J

3

918 J

4

91.8 J

Explanation:

$W = \frac{1}{2} m v_{i}^{2};$

Work, Energy and Power

268706 A body of mass $2 kg$ is projected vertically up with velocity $5 {ms}^{- 1}$ . The work done on the body by gravitational force before it is brought to rest momentarily is

1

250 J

2

25 J

3

0 J

4

- 25 J

Explanation:

$W = - m g h = - \frac{1}{2} m u^{2}$

Work, Energy and Power

268750 A body of mass $5 k g$ is moved up over $10 m$ along the line of greatest slope of a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. If $g = 10 m / s^{2}$ , the work done will be

1

500 J

2

2500 J

3

250 J

4

25 J

Explanation:

$W = \vec{F} \cdot \vec{S} = FS \cos θ$ ; here $F = mgsin α$ and
$θ = 0^{\circ}, α = 30^{\circ}$

Work, Energy and Power

268751 A particle of mass $0.5 kg$ is displaced from position $\vec{r_{1}} (2, 3, 1)$ to $\vec{r_{2}} (4, 3, 2)$ by applying a force of magnitude $30 N$ which is acting along $\exists i + \hat{j} + \hat{k} ⊟$ . The work done by the force is

1

10 \sqrt{3} J

2

30 \sqrt{3} J

3

30 J

4

40 J

Explanation:

$F_{x} = F \cos α, F_{y} = F \cos β, F_{z} = F \cos γ$ ,
$W = F . S$

Work, Energy and Power

268752 Kinetic energy of a particle moving in a straight line varies with time $t$ as $K = 4 t^{2}$ . The force acting on the particle

1 is constant

2 is increasing

3 is decreasing

4 first increases and then decreases

Explanation:

$\frac{1}{2} m v^{2} = 4 t^{2}; ∴ v = \sqrt{\frac{8}{m}} t$
comparing with $v = at, a =$ constant
i.e., the force acting on the particle is constant

Work, Energy and Power

268705 A tennis ball has a mass of $56.7 gm$ and is served by a player with a speed of $180 kmph$ . The work done in serving the ball is nearly

1

710 J

2

71 J

3

918 J

4

91.8 J

Explanation:

$W = \frac{1}{2} m v_{i}^{2};$

Work, Energy and Power

268706 A body of mass $2 kg$ is projected vertically up with velocity $5 {ms}^{- 1}$ . The work done on the body by gravitational force before it is brought to rest momentarily is

1

250 J

2

25 J

3

0 J

4

- 25 J

Explanation:

$W = - m g h = - \frac{1}{2} m u^{2}$

Work, Energy and Power

268750 A body of mass $5 k g$ is moved up over $10 m$ along the line of greatest slope of a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. If $g = 10 m / s^{2}$ , the work done will be

1

500 J

2

2500 J

3

250 J

4

25 J

Explanation:

$W = \vec{F} \cdot \vec{S} = FS \cos θ$ ; here $F = mgsin α$ and
$θ = 0^{\circ}, α = 30^{\circ}$

Work, Energy and Power

268751 A particle of mass $0.5 kg$ is displaced from position $\vec{r_{1}} (2, 3, 1)$ to $\vec{r_{2}} (4, 3, 2)$ by applying a force of magnitude $30 N$ which is acting along $\exists i + \hat{j} + \hat{k} ⊟$ . The work done by the force is

1

10 \sqrt{3} J

2

30 \sqrt{3} J

3

30 J

4

40 J

Explanation:

$F_{x} = F \cos α, F_{y} = F \cos β, F_{z} = F \cos γ$ ,
$W = F . S$

Work, Energy and Power

268752 Kinetic energy of a particle moving in a straight line varies with time $t$ as $K = 4 t^{2}$ . The force acting on the particle

1 is constant

2 is increasing

3 is decreasing

4 first increases and then decreases

Explanation:

$\frac{1}{2} m v^{2} = 4 t^{2}; ∴ v = \sqrt{\frac{8}{m}} t$
comparing with $v = at, a =$ constant
i.e., the force acting on the particle is constant

Work, Energy and Power

268705 A tennis ball has a mass of $56.7 gm$ and is served by a player with a speed of $180 kmph$ . The work done in serving the ball is nearly

1

710 J

2

71 J

3

918 J

4

91.8 J

Explanation:

$W = \frac{1}{2} m v_{i}^{2};$

Work, Energy and Power

268706 A body of mass $2 kg$ is projected vertically up with velocity $5 {ms}^{- 1}$ . The work done on the body by gravitational force before it is brought to rest momentarily is

1

250 J

2

25 J

3

0 J

4

- 25 J

Explanation:

$W = - m g h = - \frac{1}{2} m u^{2}$

Work, Energy and Power

268750 A body of mass $5 k g$ is moved up over $10 m$ along the line of greatest slope of a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. If $g = 10 m / s^{2}$ , the work done will be

1

500 J

2

2500 J

3

250 J

4

25 J

Explanation:

$W = \vec{F} \cdot \vec{S} = FS \cos θ$ ; here $F = mgsin α$ and
$θ = 0^{\circ}, α = 30^{\circ}$

Work, Energy and Power

268751 A particle of mass $0.5 kg$ is displaced from position $\vec{r_{1}} (2, 3, 1)$ to $\vec{r_{2}} (4, 3, 2)$ by applying a force of magnitude $30 N$ which is acting along $\exists i + \hat{j} + \hat{k} ⊟$ . The work done by the force is

1

10 \sqrt{3} J

2

30 \sqrt{3} J

3

30 J

4

40 J

Explanation:

$F_{x} = F \cos α, F_{y} = F \cos β, F_{z} = F \cos γ$ ,
$W = F . S$

Work, Energy and Power

268752 Kinetic energy of a particle moving in a straight line varies with time $t$ as $K = 4 t^{2}$ . The force acting on the particle

1 is constant

2 is increasing

3 is decreasing

4 first increases and then decreases

Explanation:

$\frac{1}{2} m v^{2} = 4 t^{2}; ∴ v = \sqrt{\frac{8}{m}} t$
comparing with $v = at, a =$ constant
i.e., the force acting on the particle is constant

Work, Energy and Power

268705 A tennis ball has a mass of $56.7 gm$ and is served by a player with a speed of $180 kmph$ . The work done in serving the ball is nearly

1

710 J

2

71 J

3

918 J

4

91.8 J

Explanation:

$W = \frac{1}{2} m v_{i}^{2};$

Work, Energy and Power

268706 A body of mass $2 kg$ is projected vertically up with velocity $5 {ms}^{- 1}$ . The work done on the body by gravitational force before it is brought to rest momentarily is

1

250 J

2

25 J

3

0 J

4

- 25 J

Explanation:

$W = - m g h = - \frac{1}{2} m u^{2}$

Work, Energy and Power

268750 A body of mass $5 k g$ is moved up over $10 m$ along the line of greatest slope of a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. If $g = 10 m / s^{2}$ , the work done will be

1

500 J

2

2500 J

3

250 J

4

25 J

Explanation:

$W = \vec{F} \cdot \vec{S} = FS \cos θ$ ; here $F = mgsin α$ and
$θ = 0^{\circ}, α = 30^{\circ}$

Work, Energy and Power

268751 A particle of mass $0.5 kg$ is displaced from position $\vec{r_{1}} (2, 3, 1)$ to $\vec{r_{2}} (4, 3, 2)$ by applying a force of magnitude $30 N$ which is acting along $\exists i + \hat{j} + \hat{k} ⊟$ . The work done by the force is

1

10 \sqrt{3} J

2

30 \sqrt{3} J

3

30 J

4

40 J

Explanation:

$F_{x} = F \cos α, F_{y} = F \cos β, F_{z} = F \cos γ$ ,
$W = F . S$

Work, Energy and Power

268752 Kinetic energy of a particle moving in a straight line varies with time $t$ as $K = 4 t^{2}$ . The force acting on the particle

1 is constant

2 is increasing

3 is decreasing

4 first increases and then decreases

Explanation:

$\frac{1}{2} m v^{2} = 4 t^{2}; ∴ v = \sqrt{\frac{8}{m}} t$
comparing with $v = at, a =$ constant
i.e., the force acting on the particle is constant