QBManager

Work, Energy and Power

148732 A circular ring of mass $10 kg$ and radius $1 m$ is rotating at 210 revolutions in a minute. It is brought to stop in $2 s$ . The required average power is

1

980 W

2

1210 W

3

1340 W

4

1580 W

Explanation:

B Given, mass (m) $= 10 kg$ , time $(t) = 2 \sec$
Radius of ring $(r) = 1 m$ .
$N = 210$ Revolution/min.
$ω = 2 π N$
$= 2 \times \frac{22}{7} \times \frac{210}{60}$
$= 22 rad / \sec$
$τ = I α$
$τ = {mr}^{2} α$
$τ = 10 \times (1)^{2} \frac{ω}{t} (α = \frac{ω}{t})$
$= \frac{10 \times 22}{2} = 110$
$P_{avg} = \frac{τ . ω}{t}$
$= \frac{110 \times 22}{2}$
$P_{avg} = 1210 W$

AP EAPCET-12.07.2022

Work, Energy and Power

148733 A Block of mass $50 kg$ is pulled at a constant speed of $4 m s^{- 1}$ across a horizontal floor by an applied force of $500 N$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is

1

\frac{2000}{\sqrt{3}}

2

500 \sqrt{3}

3 1732

4 1864

Explanation:

C Given, mass of block (m) $= 50 kg$
$speed (v) = 4 m / \sec, force (F) = 500 N$
$angle (θ) = 30^{\circ}$
The power associated with force $\vec{F}$ is, $P = \vec{F}$ . $\vec{v}$
And
$P = F \lor \cos θ$
$P = (500 N) \times (4.0 m / s) \cos 30^{\circ}$
$P = 2000 \times \frac{\sqrt{3}}{2}$
$P = 1000 \times 1.732$
$P = 1732$

AP EAMCET-05.07.2022

Work, Energy and Power

148734 A particle of mass $500 gm$ is moving in a straight line with velocity $v = b x^{5 / 2}$ . The work done by the net force during its displacement from $x = 0$ to $x = 4 m$ is :
$(Take b = 0.25 m^{- 3 / 2} s^{- 1})$

1

2 J

2

4 J

3

8 J

4

16 J

Explanation:

D Given that,
$Mass (m) = 500 gm = \frac{1}{2} kg, v = {bx}^{5 / 2}$
$b = 0.25 m^{- 3 / 2} s^{- 1} = \frac{1}{4} m^{- 3 / 2} s^{- 1}$
Then, $v = \frac{1}{4} x^{5 / 2} m^{- 3 / 2} s^{- 1}$
$If x = 0, v_{1} = 0$
$x = 4, v_{2} = \frac{1}{4} (4 m)^{5 / 2} m^{- 3 / 2} s^{- 1}$
$v_{2} = \frac{1}{4} \cdot {(2^{2})}^{5 / 2} m / s$
$v_{2} = 8 m / s$
$∵$ Work done $(W) =$ Change in K.E.
$= \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2}$
$= \frac{1}{2} \times \frac{1}{2} \times (8)^{2} - \frac{1}{2} \times \frac{1}{2} \times 0$
$W = 16 J$

JEE Main-29.06.2022

Work, Energy and Power

148735 Sand is being dropped from a stationary dropper at a rate of $0.5 {kgs}^{- 1}$ on a conveyor belt moving with a velocity of $5 {ms}^{- 1}$ . The power needed to keep belt moving with the same velocity will be:

1

1.25 W

2

2.5 W

3

6.25 W

4

12.5 W

Explanation:

D Mass of sand drops at a rate of $0.5 kg / s$ .
$\frac{dm}{dt} = 0.5 kg / s$
Velocity of conveyor belt $(v) = 5 m / s$
Now
Force required to keep the ball moving,
$F = \frac{d}{dt} (mv) = v \frac{dm}{dt} = 0.5 \times 5 = 2.5 N$
Then,
$Power delivered by the motor (P) = F . v$
$= 2.5 \times 5$
$= 12.5 W$

JEE Main-27.07.2022

Work, Energy and Power

148732 A circular ring of mass $10 kg$ and radius $1 m$ is rotating at 210 revolutions in a minute. It is brought to stop in $2 s$ . The required average power is

1

980 W

2

1210 W

3

1340 W

4

1580 W

Explanation:

B Given, mass (m) $= 10 kg$ , time $(t) = 2 \sec$
Radius of ring $(r) = 1 m$ .
$N = 210$ Revolution/min.
$ω = 2 π N$
$= 2 \times \frac{22}{7} \times \frac{210}{60}$
$= 22 rad / \sec$
$τ = I α$
$τ = {mr}^{2} α$
$τ = 10 \times (1)^{2} \frac{ω}{t} (α = \frac{ω}{t})$
$= \frac{10 \times 22}{2} = 110$
$P_{avg} = \frac{τ . ω}{t}$
$= \frac{110 \times 22}{2}$
$P_{avg} = 1210 W$

AP EAPCET-12.07.2022

Work, Energy and Power

148733 A Block of mass $50 kg$ is pulled at a constant speed of $4 m s^{- 1}$ across a horizontal floor by an applied force of $500 N$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is

1

\frac{2000}{\sqrt{3}}

2

500 \sqrt{3}

3 1732

4 1864

Explanation:

C Given, mass of block (m) $= 50 kg$
$speed (v) = 4 m / \sec, force (F) = 500 N$
$angle (θ) = 30^{\circ}$
The power associated with force $\vec{F}$ is, $P = \vec{F}$ . $\vec{v}$
And
$P = F \lor \cos θ$
$P = (500 N) \times (4.0 m / s) \cos 30^{\circ}$
$P = 2000 \times \frac{\sqrt{3}}{2}$
$P = 1000 \times 1.732$
$P = 1732$

AP EAMCET-05.07.2022

Work, Energy and Power

148734 A particle of mass $500 gm$ is moving in a straight line with velocity $v = b x^{5 / 2}$ . The work done by the net force during its displacement from $x = 0$ to $x = 4 m$ is :
$(Take b = 0.25 m^{- 3 / 2} s^{- 1})$

1

2 J

2

4 J

3

8 J

4

16 J

Explanation:

D Given that,
$Mass (m) = 500 gm = \frac{1}{2} kg, v = {bx}^{5 / 2}$
$b = 0.25 m^{- 3 / 2} s^{- 1} = \frac{1}{4} m^{- 3 / 2} s^{- 1}$
Then, $v = \frac{1}{4} x^{5 / 2} m^{- 3 / 2} s^{- 1}$
$If x = 0, v_{1} = 0$
$x = 4, v_{2} = \frac{1}{4} (4 m)^{5 / 2} m^{- 3 / 2} s^{- 1}$
$v_{2} = \frac{1}{4} \cdot {(2^{2})}^{5 / 2} m / s$
$v_{2} = 8 m / s$
$∵$ Work done $(W) =$ Change in K.E.
$= \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2}$
$= \frac{1}{2} \times \frac{1}{2} \times (8)^{2} - \frac{1}{2} \times \frac{1}{2} \times 0$
$W = 16 J$

JEE Main-29.06.2022

Work, Energy and Power

148735 Sand is being dropped from a stationary dropper at a rate of $0.5 {kgs}^{- 1}$ on a conveyor belt moving with a velocity of $5 {ms}^{- 1}$ . The power needed to keep belt moving with the same velocity will be:

1

1.25 W

2

2.5 W

3

6.25 W

4

12.5 W

Explanation:

D Mass of sand drops at a rate of $0.5 kg / s$ .
$\frac{dm}{dt} = 0.5 kg / s$
Velocity of conveyor belt $(v) = 5 m / s$
Now
Force required to keep the ball moving,
$F = \frac{d}{dt} (mv) = v \frac{dm}{dt} = 0.5 \times 5 = 2.5 N$
Then,
$Power delivered by the motor (P) = F . v$
$= 2.5 \times 5$
$= 12.5 W$

JEE Main-27.07.2022

Work, Energy and Power

148732 A circular ring of mass $10 kg$ and radius $1 m$ is rotating at 210 revolutions in a minute. It is brought to stop in $2 s$ . The required average power is

1

980 W

2

1210 W

3

1340 W

4

1580 W

Explanation:

B Given, mass (m) $= 10 kg$ , time $(t) = 2 \sec$
Radius of ring $(r) = 1 m$ .
$N = 210$ Revolution/min.
$ω = 2 π N$
$= 2 \times \frac{22}{7} \times \frac{210}{60}$
$= 22 rad / \sec$
$τ = I α$
$τ = {mr}^{2} α$
$τ = 10 \times (1)^{2} \frac{ω}{t} (α = \frac{ω}{t})$
$= \frac{10 \times 22}{2} = 110$
$P_{avg} = \frac{τ . ω}{t}$
$= \frac{110 \times 22}{2}$
$P_{avg} = 1210 W$

AP EAPCET-12.07.2022

Work, Energy and Power

148733 A Block of mass $50 kg$ is pulled at a constant speed of $4 m s^{- 1}$ across a horizontal floor by an applied force of $500 N$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is

1

\frac{2000}{\sqrt{3}}

2

500 \sqrt{3}

3 1732

4 1864

Explanation:

C Given, mass of block (m) $= 50 kg$
$speed (v) = 4 m / \sec, force (F) = 500 N$
$angle (θ) = 30^{\circ}$
The power associated with force $\vec{F}$ is, $P = \vec{F}$ . $\vec{v}$
And
$P = F \lor \cos θ$
$P = (500 N) \times (4.0 m / s) \cos 30^{\circ}$
$P = 2000 \times \frac{\sqrt{3}}{2}$
$P = 1000 \times 1.732$
$P = 1732$

AP EAMCET-05.07.2022

Work, Energy and Power

148734 A particle of mass $500 gm$ is moving in a straight line with velocity $v = b x^{5 / 2}$ . The work done by the net force during its displacement from $x = 0$ to $x = 4 m$ is :
$(Take b = 0.25 m^{- 3 / 2} s^{- 1})$

1

2 J

2

4 J

3

8 J

4

16 J

Explanation:

D Given that,
$Mass (m) = 500 gm = \frac{1}{2} kg, v = {bx}^{5 / 2}$
$b = 0.25 m^{- 3 / 2} s^{- 1} = \frac{1}{4} m^{- 3 / 2} s^{- 1}$
Then, $v = \frac{1}{4} x^{5 / 2} m^{- 3 / 2} s^{- 1}$
$If x = 0, v_{1} = 0$
$x = 4, v_{2} = \frac{1}{4} (4 m)^{5 / 2} m^{- 3 / 2} s^{- 1}$
$v_{2} = \frac{1}{4} \cdot {(2^{2})}^{5 / 2} m / s$
$v_{2} = 8 m / s$
$∵$ Work done $(W) =$ Change in K.E.
$= \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2}$
$= \frac{1}{2} \times \frac{1}{2} \times (8)^{2} - \frac{1}{2} \times \frac{1}{2} \times 0$
$W = 16 J$

JEE Main-29.06.2022

Work, Energy and Power

148735 Sand is being dropped from a stationary dropper at a rate of $0.5 {kgs}^{- 1}$ on a conveyor belt moving with a velocity of $5 {ms}^{- 1}$ . The power needed to keep belt moving with the same velocity will be:

1

1.25 W

2

2.5 W

3

6.25 W

4

12.5 W

Explanation:

D Mass of sand drops at a rate of $0.5 kg / s$ .
$\frac{dm}{dt} = 0.5 kg / s$
Velocity of conveyor belt $(v) = 5 m / s$
Now
Force required to keep the ball moving,
$F = \frac{d}{dt} (mv) = v \frac{dm}{dt} = 0.5 \times 5 = 2.5 N$
Then,
$Power delivered by the motor (P) = F . v$
$= 2.5 \times 5$
$= 12.5 W$

JEE Main-27.07.2022

Work, Energy and Power

148732 A circular ring of mass $10 kg$ and radius $1 m$ is rotating at 210 revolutions in a minute. It is brought to stop in $2 s$ . The required average power is

1

980 W

2

1210 W

3

1340 W

4

1580 W

Explanation:

B Given, mass (m) $= 10 kg$ , time $(t) = 2 \sec$
Radius of ring $(r) = 1 m$ .
$N = 210$ Revolution/min.
$ω = 2 π N$
$= 2 \times \frac{22}{7} \times \frac{210}{60}$
$= 22 rad / \sec$
$τ = I α$
$τ = {mr}^{2} α$
$τ = 10 \times (1)^{2} \frac{ω}{t} (α = \frac{ω}{t})$
$= \frac{10 \times 22}{2} = 110$
$P_{avg} = \frac{τ . ω}{t}$
$= \frac{110 \times 22}{2}$
$P_{avg} = 1210 W$

AP EAPCET-12.07.2022

Work, Energy and Power

148733 A Block of mass $50 kg$ is pulled at a constant speed of $4 m s^{- 1}$ across a horizontal floor by an applied force of $500 N$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is

1

\frac{2000}{\sqrt{3}}

2

500 \sqrt{3}

3 1732

4 1864

Explanation:

C Given, mass of block (m) $= 50 kg$
$speed (v) = 4 m / \sec, force (F) = 500 N$
$angle (θ) = 30^{\circ}$
The power associated with force $\vec{F}$ is, $P = \vec{F}$ . $\vec{v}$
And
$P = F \lor \cos θ$
$P = (500 N) \times (4.0 m / s) \cos 30^{\circ}$
$P = 2000 \times \frac{\sqrt{3}}{2}$
$P = 1000 \times 1.732$
$P = 1732$

AP EAMCET-05.07.2022

Work, Energy and Power

148734 A particle of mass $500 gm$ is moving in a straight line with velocity $v = b x^{5 / 2}$ . The work done by the net force during its displacement from $x = 0$ to $x = 4 m$ is :
$(Take b = 0.25 m^{- 3 / 2} s^{- 1})$

1

2 J

2

4 J

3

8 J

4

16 J

Explanation:

D Given that,
$Mass (m) = 500 gm = \frac{1}{2} kg, v = {bx}^{5 / 2}$
$b = 0.25 m^{- 3 / 2} s^{- 1} = \frac{1}{4} m^{- 3 / 2} s^{- 1}$
Then, $v = \frac{1}{4} x^{5 / 2} m^{- 3 / 2} s^{- 1}$
$If x = 0, v_{1} = 0$
$x = 4, v_{2} = \frac{1}{4} (4 m)^{5 / 2} m^{- 3 / 2} s^{- 1}$
$v_{2} = \frac{1}{4} \cdot {(2^{2})}^{5 / 2} m / s$
$v_{2} = 8 m / s$
$∵$ Work done $(W) =$ Change in K.E.
$= \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2}$
$= \frac{1}{2} \times \frac{1}{2} \times (8)^{2} - \frac{1}{2} \times \frac{1}{2} \times 0$
$W = 16 J$

JEE Main-29.06.2022

Work, Energy and Power

148735 Sand is being dropped from a stationary dropper at a rate of $0.5 {kgs}^{- 1}$ on a conveyor belt moving with a velocity of $5 {ms}^{- 1}$ . The power needed to keep belt moving with the same velocity will be:

1

1.25 W

2

2.5 W

3

6.25 W

4

12.5 W

Explanation:

D Mass of sand drops at a rate of $0.5 kg / s$ .
$\frac{dm}{dt} = 0.5 kg / s$
Velocity of conveyor belt $(v) = 5 m / s$
Now
Force required to keep the ball moving,
$F = \frac{d}{dt} (mv) = v \frac{dm}{dt} = 0.5 \times 5 = 2.5 N$
Then,
$Power delivered by the motor (P) = F . v$
$= 2.5 \times 5$
$= 12.5 W$

JEE Main-27.07.2022

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