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PHXI06:WORK ENERGY AND POWER

355766 The velocity-time graph of a particle, moving in a straight line, is shown in the figure. The mass of the particle is $2 k g$ . The work done by all the forces acting on the particle in time interval between $t = 0 s$ to $t = 10 s$ is
supporting img

1

300 J

2

- 300 J

3

400 J

4

- 400 J

Explanation:

From work-energy theorem,
$W = Δ K \cdot E$
$= K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 (400 - 100) = 300 J$

PHXI06:WORK ENERGY AND POWER

355767 Consider a drop of rain water having mass 1 $g$ falling from a height of 1 $k m$ . It hits the ground with a speed of 50 $m / s$ . Take ' $g$ ' constant with a value $10 m / s^{2}$ . The work done by the
(i) gravitational force and the
(ii) resistive force of air is

1

(i) 1.25 J (i i) - 8.25 J

2

(i) 100 J (i i) 8.75 J

3

(i) 10 J (i i) - 8.75 J

4

(i) - 10 J (i i) - 8.25 J

Explanation:

Work done by the gravity $(W_{g}) = m g h$
$= 10^{- 3} \times 10 \times 10^{3} = 10 J$
By work-energy theorem $= W_{g} + W_{res} = Δ K E$
$\begin{aligned} 10 + W_{res} = \frac{1}{2} \times 10^{- 3} \times (50)^{2} \\ W_{res} = - 8.75 J \end{aligned}$

NEET - 2017

PHXI06:WORK ENERGY AND POWER

355768 A mass $M$ is lowered with the help of a string by a distance $h$ at a constant acceleration $g / 2$ . The work done by the string will be

1

\frac{- 3 M g h}{2}

2

\frac{M g h}{2}

3

\frac{- M g h}{2}

4

\frac{3 M g h}{2}

Explanation:

Tension in the string
$\begin{aligned} T & = M (g - a) = M (g - \frac{g}{2}) = \frac{M g}{2} \\ W & = Force \times displacement \end{aligned}$
$= - \frac{M g h}{2}$

PHXI06:WORK ENERGY AND POWER

355769 A man is rowing a boat upstream and inspite of that the boat is found to be not moving with respect to the bank. The work done by the man is

1 Zero

2 Positive

3 Negative

4 May be

+ v e

or

- v e

PHXI06:WORK ENERGY AND POWER

355766 The velocity-time graph of a particle, moving in a straight line, is shown in the figure. The mass of the particle is $2 k g$ . The work done by all the forces acting on the particle in time interval between $t = 0 s$ to $t = 10 s$ is
supporting img

1

300 J

2

- 300 J

3

400 J

4

- 400 J

Explanation:

From work-energy theorem,
$W = Δ K \cdot E$
$= K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 (400 - 100) = 300 J$

PHXI06:WORK ENERGY AND POWER

355767 Consider a drop of rain water having mass 1 $g$ falling from a height of 1 $k m$ . It hits the ground with a speed of 50 $m / s$ . Take ' $g$ ' constant with a value $10 m / s^{2}$ . The work done by the
(i) gravitational force and the
(ii) resistive force of air is

1

(i) 1.25 J (i i) - 8.25 J

2

(i) 100 J (i i) 8.75 J

3

(i) 10 J (i i) - 8.75 J

4

(i) - 10 J (i i) - 8.25 J

Explanation:

Work done by the gravity $(W_{g}) = m g h$
$= 10^{- 3} \times 10 \times 10^{3} = 10 J$
By work-energy theorem $= W_{g} + W_{res} = Δ K E$
$\begin{aligned} 10 + W_{res} = \frac{1}{2} \times 10^{- 3} \times (50)^{2} \\ W_{res} = - 8.75 J \end{aligned}$

NEET - 2017

PHXI06:WORK ENERGY AND POWER

355768 A mass $M$ is lowered with the help of a string by a distance $h$ at a constant acceleration $g / 2$ . The work done by the string will be

1

\frac{- 3 M g h}{2}

2

\frac{M g h}{2}

3

\frac{- M g h}{2}

4

\frac{3 M g h}{2}

Explanation:

Tension in the string
$\begin{aligned} T & = M (g - a) = M (g - \frac{g}{2}) = \frac{M g}{2} \\ W & = Force \times displacement \end{aligned}$
$= - \frac{M g h}{2}$

PHXI06:WORK ENERGY AND POWER

355769 A man is rowing a boat upstream and inspite of that the boat is found to be not moving with respect to the bank. The work done by the man is

1 Zero

2 Positive

3 Negative

4 May be

+ v e

or

- v e

PHXI06:WORK ENERGY AND POWER

355766 The velocity-time graph of a particle, moving in a straight line, is shown in the figure. The mass of the particle is $2 k g$ . The work done by all the forces acting on the particle in time interval between $t = 0 s$ to $t = 10 s$ is
supporting img

1

300 J

2

- 300 J

3

400 J

4

- 400 J

Explanation:

From work-energy theorem,
$W = Δ K \cdot E$
$= K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 (400 - 100) = 300 J$

PHXI06:WORK ENERGY AND POWER

355767 Consider a drop of rain water having mass 1 $g$ falling from a height of 1 $k m$ . It hits the ground with a speed of 50 $m / s$ . Take ' $g$ ' constant with a value $10 m / s^{2}$ . The work done by the
(i) gravitational force and the
(ii) resistive force of air is

1

(i) 1.25 J (i i) - 8.25 J

2

(i) 100 J (i i) 8.75 J

3

(i) 10 J (i i) - 8.75 J

4

(i) - 10 J (i i) - 8.25 J

Explanation:

Work done by the gravity $(W_{g}) = m g h$
$= 10^{- 3} \times 10 \times 10^{3} = 10 J$
By work-energy theorem $= W_{g} + W_{res} = Δ K E$
$\begin{aligned} 10 + W_{res} = \frac{1}{2} \times 10^{- 3} \times (50)^{2} \\ W_{res} = - 8.75 J \end{aligned}$

NEET - 2017

PHXI06:WORK ENERGY AND POWER

355768 A mass $M$ is lowered with the help of a string by a distance $h$ at a constant acceleration $g / 2$ . The work done by the string will be

1

\frac{- 3 M g h}{2}

2

\frac{M g h}{2}

3

\frac{- M g h}{2}

4

\frac{3 M g h}{2}

Explanation:

Tension in the string
$\begin{aligned} T & = M (g - a) = M (g - \frac{g}{2}) = \frac{M g}{2} \\ W & = Force \times displacement \end{aligned}$
$= - \frac{M g h}{2}$

PHXI06:WORK ENERGY AND POWER

355769 A man is rowing a boat upstream and inspite of that the boat is found to be not moving with respect to the bank. The work done by the man is

1 Zero

2 Positive

3 Negative

4 May be

+ v e

or

- v e

PHXI06:WORK ENERGY AND POWER

355766 The velocity-time graph of a particle, moving in a straight line, is shown in the figure. The mass of the particle is $2 k g$ . The work done by all the forces acting on the particle in time interval between $t = 0 s$ to $t = 10 s$ is
supporting img

1

300 J

2

- 300 J

3

400 J

4

- 400 J

Explanation:

From work-energy theorem,
$W = Δ K \cdot E$
$= K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 (400 - 100) = 300 J$

PHXI06:WORK ENERGY AND POWER

355767 Consider a drop of rain water having mass 1 $g$ falling from a height of 1 $k m$ . It hits the ground with a speed of 50 $m / s$ . Take ' $g$ ' constant with a value $10 m / s^{2}$ . The work done by the
(i) gravitational force and the
(ii) resistive force of air is

1

(i) 1.25 J (i i) - 8.25 J

2

(i) 100 J (i i) 8.75 J

3

(i) 10 J (i i) - 8.75 J

4

(i) - 10 J (i i) - 8.25 J

Explanation:

Work done by the gravity $(W_{g}) = m g h$
$= 10^{- 3} \times 10 \times 10^{3} = 10 J$
By work-energy theorem $= W_{g} + W_{res} = Δ K E$
$\begin{aligned} 10 + W_{res} = \frac{1}{2} \times 10^{- 3} \times (50)^{2} \\ W_{res} = - 8.75 J \end{aligned}$

NEET - 2017

PHXI06:WORK ENERGY AND POWER

355768 A mass $M$ is lowered with the help of a string by a distance $h$ at a constant acceleration $g / 2$ . The work done by the string will be

1

\frac{- 3 M g h}{2}

2

\frac{M g h}{2}

3

\frac{- M g h}{2}

4

\frac{3 M g h}{2}

Explanation:

Tension in the string
$\begin{aligned} T & = M (g - a) = M (g - \frac{g}{2}) = \frac{M g}{2} \\ W & = Force \times displacement \end{aligned}$
$= - \frac{M g h}{2}$

PHXI06:WORK ENERGY AND POWER

355769 A man is rowing a boat upstream and inspite of that the boat is found to be not moving with respect to the bank. The work done by the man is

1 Zero

2 Positive

3 Negative

4 May be

+ v e

or

- v e