QBManager

Work, Energy and Power

148925 If the linear momentum is increased by $50 %$ , the kinetic energy will increase by:

1

50 %

2

100 %

3

125 %

4

10 %

Explanation:

C Given, linear momentum increased $= 50 %$ .
We know that,
Initial kinetic energy (K.E. $)_{i} = \frac{p_{i}^{2}}{2 m}$
Final kinetic energy $(K.E.)_{f} = \frac{{(1.5 p_{i})}^{2}}{2 m}$
or $(K.E.)_{f} = \frac{2.25 P_{i}^{2}}{2 m} = 2.25 (K.E.)_{i}$
Increase kinetic energy $= \frac{(K.E.)_{f} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= \frac{2.25 (K.E.)_{i} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= 125 %$
Hence, change in K.E. $= 125 %$ of initial

AIIMS-2013

Work, Energy and Power

148926 A bullet of mass $20 g$ and moving with $600 m / s$ collides with a block of mass $4 kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 $m$ after collision?

1

200 m / s

2

150 m / s

3

400 m / s

4

300 m / s

Explanation:

A Given,
Mass of bullet $(m_{1}) = 20 g = 20 \times 10^{- 3} kg$
Velocity of bullet before collision $(v_{1}) = 600 m / \sec$
Mass of block $(m_{2}) = 4 kg$
$h = 0.2 m$
According to law of conservation of linear momentum, $m_{1} v_{1} = m_{1} v + m_{2} v_{2}$
$v =$ velocity of bullet after collision
$v_{2} =$ velocity of block
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + 4 v_{2}$
After collision, block rises to height $0.2 m$
From the law of conservation of energy,
$m_{2} gh = \frac{1}{2} m_{2} v_{2}^{2}$
$2 gh = v_{2}^{2}$
$v_{2} = \sqrt{2 gh}$
$v_{2} = \sqrt{2 \times 10 \times 0.2} [∵ h = 0.2 m]$
$v_{2} = \sqrt{4}$
$v_{2} = 2 m / s$
Put the value $v_{2}$ in equation (i),
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + (4 \times 2)$
$2 \times 6 = (2 \times 10^{- 2}) v + 8$
$(2 \times 10^{- 2}) v = 12 - 8$
$(2 \times 10^{- 2}) v = 4$
$v = \frac{4}{(2 \times 10^{- 2})}$
$v = 2 \times 10^{2}$
$v = 200 m / \sec$

AIIMS-27.05.2018(M)]

Work, Energy and Power

148927 The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 kg$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

1

10 J

2

20 J

3

40 J

4

50 J

Explanation:

A Given,
$\begin{array}{r} Mass of the body (m) = 25 kg \\ Velocity (v) = 2 m / s \end{array}$
Initial kinetic energy (K.E. $)_{1} = \frac{1}{2} {mv}^{2}$
$(K.E.)_{1} = \frac{1}{2} \times 25 \times 2 \times 2$
$(K.E.)_{1} = 50 J$
When the distance covered by the body is $4 m$ , its kinetic energy will be reduced due to resistive force because body will work against this resistive force.
Therefore,
Work done against resistive force
$=$ Area under $F - x$ graph $=$ Area of triangle
$= \frac{1}{2} \times base \times height$
$= \frac{1}{2} \times 4 \times 20 = 40 J$
Final kinetic energy
$=$ initial kinetic energy - work done against resistive force
$= 50 - 40$
$= 10 J$

AP EAMCET-03.09.2021 Shift-I]

Work, Energy and Power

148928 If potential energy is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ . Then find out maximum force. (Given $a = 2, b = 4$ )

1

- \frac{16}{27} N

2

- \frac{32}{27} N

3

+ \frac{32}{27} N

4

+ \frac{16}{27} N

Explanation:

A Given,
The potential energy,
$U = \frac{a}{r^{2}} - \frac{b}{r}$
And, $a = 2, b = 4$
We know that,
Force on the system will be given by
$F = - \frac{d U}{d r}$
$F = - \frac{d}{d r} [\frac{a}{r^{2}} - \frac{b}{r}]$
$F = - \frac{d}{d r} [a^{- 2} - b r^{- 1}]$
$F = - a \frac{d}{d r} r^{- 2} + b \frac{d}{d r} r^{- 1}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Force will be maximum if, $\frac{dF}{dr} = 0$
$\frac{dF}{dr} = 0$
$\Rightarrow \frac{d}{dr} [\frac{2 a}{r^{3}} - \frac{b}{r^{2}}] = 0$
$\Rightarrow 2 a \frac{d}{dr} r^{- 3} - b \frac{d}{dr} r^{- 2} = 0$
$\Rightarrow - 6 {ar}^{- 4} + 2 {br}^{- 3} = 0$
$\Rightarrow - \frac{6 a}{r^{4}} + \frac{2 b}{r^{3}} = 0$
$\Rightarrow \frac{3 a}{r} = 3 a$
$\Rightarrow r = 3 (\frac{a}{b})$
Put the value $(a = 2)$ and $(b = 4)$ in above equation
$\Rightarrow r = 3 (\frac{2}{4})$
$\Rightarrow r = \frac{3}{2}$
Taking equation (i)
$F_{max} = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Put the value $a, b$ and $r$ in above equation.
$F_{max} = \frac{2 \times 2}{{(\frac{3}{2})}^{3}} - \frac{4}{{(\frac{3}{2})}^{2}}$
$F_{max} = \frac{4}{\frac{27}{8}} - \frac{4}{\frac{9}{4}}$
$F_{max} = \frac{4 \times 8}{27} - \frac{4 \times 4}{9}$
$F_{max} = \frac{16}{9} (\frac{2}{3} - 1)$
$F_{max} = - \frac{16}{9 \times 3}$
$F_{max} = - \frac{16}{27} N$

TS EAMCET-04.08.2021 Shift-I]

Work, Energy and Power

148929 A ball of mass $2 kg$ is thrown from a tall building with velocity,
$v = (20 m / s) \hat{i} + (24 m / s) \hat{j}$ at time $t = 0 s$ .
Change in the potential energy of the ball after, $t = 8 s$ is (The ball is assumed to be in air during its motion between $0 s$ and $8 s$ , $\hat{i}$ is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g = 10 m / s^{2}$ )

1

- 2.56 kJ

2

0.52 kJ

3

1.76 kJ

4

- 2.44 kJ

Explanation:

A Given that,
Initial velocity of ball $(u) = (20 \hat{i} + 24 \hat{j}) m / s$ at time $t = 0$ $\sec$
$Mass (m) = 2 kg$
Let $v =$ velocity at time $t = 8 \sec$
Then after time $t = 8 \sec$
Let the equation of motion be,
$v = u - gt$
$v = (20 \hat{i} + 24 \hat{j}) - (10 \hat{j}) \times 8$
$(∵ g always work in vertical direction)$
$v = (20 \hat{i} - 56 \hat{j}) m / s$
From the law of conservation of energy,
Change in potential energy $=$ Change in kinetic energy
$Δ P.E. = Δ K.E.$
$Δ$ P.E. $= \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$
$Δ P.E. = \frac{1}{2} m [u . u - v . v] = \frac{1}{2} \times 2 [u . u - v . v]$
$Δ$ P.E. $= [(20 \hat{i} + 24 \hat{j}) \cdot (20 \hat{i} + 24 \hat{j})] - [(20 \hat{i} - 56 \hat{j}) \cdot (20 \hat{i} - 56 \hat{j})]$
$Δ P.E. = [(20)^{2} + (24)^{2}] - [(20)^{2} + (56)^{2}]$
$Δ$ P.E. $= 576 - 3136$
$Δ$ P.E. $= - 2560 J = - 2.56 kJ$

TS- EAMCET-10.09.2020

Work, Energy and Power

148925 If the linear momentum is increased by $50 %$ , the kinetic energy will increase by:

1

50 %

2

100 %

3

125 %

4

10 %

Explanation:

C Given, linear momentum increased $= 50 %$ .
We know that,
Initial kinetic energy (K.E. $)_{i} = \frac{p_{i}^{2}}{2 m}$
Final kinetic energy $(K.E.)_{f} = \frac{{(1.5 p_{i})}^{2}}{2 m}$
or $(K.E.)_{f} = \frac{2.25 P_{i}^{2}}{2 m} = 2.25 (K.E.)_{i}$
Increase kinetic energy $= \frac{(K.E.)_{f} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= \frac{2.25 (K.E.)_{i} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= 125 %$
Hence, change in K.E. $= 125 %$ of initial

AIIMS-2013

Work, Energy and Power

148926 A bullet of mass $20 g$ and moving with $600 m / s$ collides with a block of mass $4 kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 $m$ after collision?

1

200 m / s

2

150 m / s

3

400 m / s

4

300 m / s

Explanation:

A Given,
Mass of bullet $(m_{1}) = 20 g = 20 \times 10^{- 3} kg$
Velocity of bullet before collision $(v_{1}) = 600 m / \sec$
Mass of block $(m_{2}) = 4 kg$
$h = 0.2 m$
According to law of conservation of linear momentum, $m_{1} v_{1} = m_{1} v + m_{2} v_{2}$
$v =$ velocity of bullet after collision
$v_{2} =$ velocity of block
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + 4 v_{2}$
After collision, block rises to height $0.2 m$
From the law of conservation of energy,
$m_{2} gh = \frac{1}{2} m_{2} v_{2}^{2}$
$2 gh = v_{2}^{2}$
$v_{2} = \sqrt{2 gh}$
$v_{2} = \sqrt{2 \times 10 \times 0.2} [∵ h = 0.2 m]$
$v_{2} = \sqrt{4}$
$v_{2} = 2 m / s$
Put the value $v_{2}$ in equation (i),
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + (4 \times 2)$
$2 \times 6 = (2 \times 10^{- 2}) v + 8$
$(2 \times 10^{- 2}) v = 12 - 8$
$(2 \times 10^{- 2}) v = 4$
$v = \frac{4}{(2 \times 10^{- 2})}$
$v = 2 \times 10^{2}$
$v = 200 m / \sec$

AIIMS-27.05.2018(M)]

Work, Energy and Power

148927 The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 kg$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

1

10 J

2

20 J

3

40 J

4

50 J

Explanation:

A Given,
$\begin{array}{r} Mass of the body (m) = 25 kg \\ Velocity (v) = 2 m / s \end{array}$
Initial kinetic energy (K.E. $)_{1} = \frac{1}{2} {mv}^{2}$
$(K.E.)_{1} = \frac{1}{2} \times 25 \times 2 \times 2$
$(K.E.)_{1} = 50 J$
When the distance covered by the body is $4 m$ , its kinetic energy will be reduced due to resistive force because body will work against this resistive force.
Therefore,
Work done against resistive force
$=$ Area under $F - x$ graph $=$ Area of triangle
$= \frac{1}{2} \times base \times height$
$= \frac{1}{2} \times 4 \times 20 = 40 J$
Final kinetic energy
$=$ initial kinetic energy - work done against resistive force
$= 50 - 40$
$= 10 J$

AP EAMCET-03.09.2021 Shift-I]

Work, Energy and Power

148928 If potential energy is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ . Then find out maximum force. (Given $a = 2, b = 4$ )

1

- \frac{16}{27} N

2

- \frac{32}{27} N

3

+ \frac{32}{27} N

4

+ \frac{16}{27} N

Explanation:

A Given,
The potential energy,
$U = \frac{a}{r^{2}} - \frac{b}{r}$
And, $a = 2, b = 4$
We know that,
Force on the system will be given by
$F = - \frac{d U}{d r}$
$F = - \frac{d}{d r} [\frac{a}{r^{2}} - \frac{b}{r}]$
$F = - \frac{d}{d r} [a^{- 2} - b r^{- 1}]$
$F = - a \frac{d}{d r} r^{- 2} + b \frac{d}{d r} r^{- 1}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Force will be maximum if, $\frac{dF}{dr} = 0$
$\frac{dF}{dr} = 0$
$\Rightarrow \frac{d}{dr} [\frac{2 a}{r^{3}} - \frac{b}{r^{2}}] = 0$
$\Rightarrow 2 a \frac{d}{dr} r^{- 3} - b \frac{d}{dr} r^{- 2} = 0$
$\Rightarrow - 6 {ar}^{- 4} + 2 {br}^{- 3} = 0$
$\Rightarrow - \frac{6 a}{r^{4}} + \frac{2 b}{r^{3}} = 0$
$\Rightarrow \frac{3 a}{r} = 3 a$
$\Rightarrow r = 3 (\frac{a}{b})$
Put the value $(a = 2)$ and $(b = 4)$ in above equation
$\Rightarrow r = 3 (\frac{2}{4})$
$\Rightarrow r = \frac{3}{2}$
Taking equation (i)
$F_{max} = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Put the value $a, b$ and $r$ in above equation.
$F_{max} = \frac{2 \times 2}{{(\frac{3}{2})}^{3}} - \frac{4}{{(\frac{3}{2})}^{2}}$
$F_{max} = \frac{4}{\frac{27}{8}} - \frac{4}{\frac{9}{4}}$
$F_{max} = \frac{4 \times 8}{27} - \frac{4 \times 4}{9}$
$F_{max} = \frac{16}{9} (\frac{2}{3} - 1)$
$F_{max} = - \frac{16}{9 \times 3}$
$F_{max} = - \frac{16}{27} N$

TS EAMCET-04.08.2021 Shift-I]

Work, Energy and Power

148929 A ball of mass $2 kg$ is thrown from a tall building with velocity,
$v = (20 m / s) \hat{i} + (24 m / s) \hat{j}$ at time $t = 0 s$ .
Change in the potential energy of the ball after, $t = 8 s$ is (The ball is assumed to be in air during its motion between $0 s$ and $8 s$ , $\hat{i}$ is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g = 10 m / s^{2}$ )

1

- 2.56 kJ

2

0.52 kJ

3

1.76 kJ

4

- 2.44 kJ

Explanation:

A Given that,
Initial velocity of ball $(u) = (20 \hat{i} + 24 \hat{j}) m / s$ at time $t = 0$ $\sec$
$Mass (m) = 2 kg$
Let $v =$ velocity at time $t = 8 \sec$
Then after time $t = 8 \sec$
Let the equation of motion be,
$v = u - gt$
$v = (20 \hat{i} + 24 \hat{j}) - (10 \hat{j}) \times 8$
$(∵ g always work in vertical direction)$
$v = (20 \hat{i} - 56 \hat{j}) m / s$
From the law of conservation of energy,
Change in potential energy $=$ Change in kinetic energy
$Δ P.E. = Δ K.E.$
$Δ$ P.E. $= \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$
$Δ P.E. = \frac{1}{2} m [u . u - v . v] = \frac{1}{2} \times 2 [u . u - v . v]$
$Δ$ P.E. $= [(20 \hat{i} + 24 \hat{j}) \cdot (20 \hat{i} + 24 \hat{j})] - [(20 \hat{i} - 56 \hat{j}) \cdot (20 \hat{i} - 56 \hat{j})]$
$Δ P.E. = [(20)^{2} + (24)^{2}] - [(20)^{2} + (56)^{2}]$
$Δ$ P.E. $= 576 - 3136$
$Δ$ P.E. $= - 2560 J = - 2.56 kJ$

TS- EAMCET-10.09.2020

Work, Energy and Power

148925 If the linear momentum is increased by $50 %$ , the kinetic energy will increase by:

1

50 %

2

100 %

3

125 %

4

10 %

Explanation:

C Given, linear momentum increased $= 50 %$ .
We know that,
Initial kinetic energy (K.E. $)_{i} = \frac{p_{i}^{2}}{2 m}$
Final kinetic energy $(K.E.)_{f} = \frac{{(1.5 p_{i})}^{2}}{2 m}$
or $(K.E.)_{f} = \frac{2.25 P_{i}^{2}}{2 m} = 2.25 (K.E.)_{i}$
Increase kinetic energy $= \frac{(K.E.)_{f} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= \frac{2.25 (K.E.)_{i} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= 125 %$
Hence, change in K.E. $= 125 %$ of initial

AIIMS-2013

Work, Energy and Power

148926 A bullet of mass $20 g$ and moving with $600 m / s$ collides with a block of mass $4 kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 $m$ after collision?

1

200 m / s

2

150 m / s

3

400 m / s

4

300 m / s

Explanation:

A Given,
Mass of bullet $(m_{1}) = 20 g = 20 \times 10^{- 3} kg$
Velocity of bullet before collision $(v_{1}) = 600 m / \sec$
Mass of block $(m_{2}) = 4 kg$
$h = 0.2 m$
According to law of conservation of linear momentum, $m_{1} v_{1} = m_{1} v + m_{2} v_{2}$
$v =$ velocity of bullet after collision
$v_{2} =$ velocity of block
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + 4 v_{2}$
After collision, block rises to height $0.2 m$
From the law of conservation of energy,
$m_{2} gh = \frac{1}{2} m_{2} v_{2}^{2}$
$2 gh = v_{2}^{2}$
$v_{2} = \sqrt{2 gh}$
$v_{2} = \sqrt{2 \times 10 \times 0.2} [∵ h = 0.2 m]$
$v_{2} = \sqrt{4}$
$v_{2} = 2 m / s$
Put the value $v_{2}$ in equation (i),
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + (4 \times 2)$
$2 \times 6 = (2 \times 10^{- 2}) v + 8$
$(2 \times 10^{- 2}) v = 12 - 8$
$(2 \times 10^{- 2}) v = 4$
$v = \frac{4}{(2 \times 10^{- 2})}$
$v = 2 \times 10^{2}$
$v = 200 m / \sec$

AIIMS-27.05.2018(M)]

Work, Energy and Power

148927 The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 kg$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

1

10 J

2

20 J

3

40 J

4

50 J

Explanation:

A Given,
$\begin{array}{r} Mass of the body (m) = 25 kg \\ Velocity (v) = 2 m / s \end{array}$
Initial kinetic energy (K.E. $)_{1} = \frac{1}{2} {mv}^{2}$
$(K.E.)_{1} = \frac{1}{2} \times 25 \times 2 \times 2$
$(K.E.)_{1} = 50 J$
When the distance covered by the body is $4 m$ , its kinetic energy will be reduced due to resistive force because body will work against this resistive force.
Therefore,
Work done against resistive force
$=$ Area under $F - x$ graph $=$ Area of triangle
$= \frac{1}{2} \times base \times height$
$= \frac{1}{2} \times 4 \times 20 = 40 J$
Final kinetic energy
$=$ initial kinetic energy - work done against resistive force
$= 50 - 40$
$= 10 J$

AP EAMCET-03.09.2021 Shift-I]

Work, Energy and Power

148928 If potential energy is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ . Then find out maximum force. (Given $a = 2, b = 4$ )

1

- \frac{16}{27} N

2

- \frac{32}{27} N

3

+ \frac{32}{27} N

4

+ \frac{16}{27} N

Explanation:

A Given,
The potential energy,
$U = \frac{a}{r^{2}} - \frac{b}{r}$
And, $a = 2, b = 4$
We know that,
Force on the system will be given by
$F = - \frac{d U}{d r}$
$F = - \frac{d}{d r} [\frac{a}{r^{2}} - \frac{b}{r}]$
$F = - \frac{d}{d r} [a^{- 2} - b r^{- 1}]$
$F = - a \frac{d}{d r} r^{- 2} + b \frac{d}{d r} r^{- 1}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Force will be maximum if, $\frac{dF}{dr} = 0$
$\frac{dF}{dr} = 0$
$\Rightarrow \frac{d}{dr} [\frac{2 a}{r^{3}} - \frac{b}{r^{2}}] = 0$
$\Rightarrow 2 a \frac{d}{dr} r^{- 3} - b \frac{d}{dr} r^{- 2} = 0$
$\Rightarrow - 6 {ar}^{- 4} + 2 {br}^{- 3} = 0$
$\Rightarrow - \frac{6 a}{r^{4}} + \frac{2 b}{r^{3}} = 0$
$\Rightarrow \frac{3 a}{r} = 3 a$
$\Rightarrow r = 3 (\frac{a}{b})$
Put the value $(a = 2)$ and $(b = 4)$ in above equation
$\Rightarrow r = 3 (\frac{2}{4})$
$\Rightarrow r = \frac{3}{2}$
Taking equation (i)
$F_{max} = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Put the value $a, b$ and $r$ in above equation.
$F_{max} = \frac{2 \times 2}{{(\frac{3}{2})}^{3}} - \frac{4}{{(\frac{3}{2})}^{2}}$
$F_{max} = \frac{4}{\frac{27}{8}} - \frac{4}{\frac{9}{4}}$
$F_{max} = \frac{4 \times 8}{27} - \frac{4 \times 4}{9}$
$F_{max} = \frac{16}{9} (\frac{2}{3} - 1)$
$F_{max} = - \frac{16}{9 \times 3}$
$F_{max} = - \frac{16}{27} N$

TS EAMCET-04.08.2021 Shift-I]

Work, Energy and Power

148929 A ball of mass $2 kg$ is thrown from a tall building with velocity,
$v = (20 m / s) \hat{i} + (24 m / s) \hat{j}$ at time $t = 0 s$ .
Change in the potential energy of the ball after, $t = 8 s$ is (The ball is assumed to be in air during its motion between $0 s$ and $8 s$ , $\hat{i}$ is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g = 10 m / s^{2}$ )

1

- 2.56 kJ

2

0.52 kJ

3

1.76 kJ

4

- 2.44 kJ

Explanation:

A Given that,
Initial velocity of ball $(u) = (20 \hat{i} + 24 \hat{j}) m / s$ at time $t = 0$ $\sec$
$Mass (m) = 2 kg$
Let $v =$ velocity at time $t = 8 \sec$
Then after time $t = 8 \sec$
Let the equation of motion be,
$v = u - gt$
$v = (20 \hat{i} + 24 \hat{j}) - (10 \hat{j}) \times 8$
$(∵ g always work in vertical direction)$
$v = (20 \hat{i} - 56 \hat{j}) m / s$
From the law of conservation of energy,
Change in potential energy $=$ Change in kinetic energy
$Δ P.E. = Δ K.E.$
$Δ$ P.E. $= \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$
$Δ P.E. = \frac{1}{2} m [u . u - v . v] = \frac{1}{2} \times 2 [u . u - v . v]$
$Δ$ P.E. $= [(20 \hat{i} + 24 \hat{j}) \cdot (20 \hat{i} + 24 \hat{j})] - [(20 \hat{i} - 56 \hat{j}) \cdot (20 \hat{i} - 56 \hat{j})]$
$Δ P.E. = [(20)^{2} + (24)^{2}] - [(20)^{2} + (56)^{2}]$
$Δ$ P.E. $= 576 - 3136$
$Δ$ P.E. $= - 2560 J = - 2.56 kJ$

TS- EAMCET-10.09.2020

Work, Energy and Power

148925 If the linear momentum is increased by $50 %$ , the kinetic energy will increase by:

1

50 %

2

100 %

3

125 %

4

10 %

Explanation:

C Given, linear momentum increased $= 50 %$ .
We know that,
Initial kinetic energy (K.E. $)_{i} = \frac{p_{i}^{2}}{2 m}$
Final kinetic energy $(K.E.)_{f} = \frac{{(1.5 p_{i})}^{2}}{2 m}$
or $(K.E.)_{f} = \frac{2.25 P_{i}^{2}}{2 m} = 2.25 (K.E.)_{i}$
Increase kinetic energy $= \frac{(K.E.)_{f} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= \frac{2.25 (K.E.)_{i} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= 125 %$
Hence, change in K.E. $= 125 %$ of initial

AIIMS-2013

Work, Energy and Power

148926 A bullet of mass $20 g$ and moving with $600 m / s$ collides with a block of mass $4 kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 $m$ after collision?

1

200 m / s

2

150 m / s

3

400 m / s

4

300 m / s

Explanation:

A Given,
Mass of bullet $(m_{1}) = 20 g = 20 \times 10^{- 3} kg$
Velocity of bullet before collision $(v_{1}) = 600 m / \sec$
Mass of block $(m_{2}) = 4 kg$
$h = 0.2 m$
According to law of conservation of linear momentum, $m_{1} v_{1} = m_{1} v + m_{2} v_{2}$
$v =$ velocity of bullet after collision
$v_{2} =$ velocity of block
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + 4 v_{2}$
After collision, block rises to height $0.2 m$
From the law of conservation of energy,
$m_{2} gh = \frac{1}{2} m_{2} v_{2}^{2}$
$2 gh = v_{2}^{2}$
$v_{2} = \sqrt{2 gh}$
$v_{2} = \sqrt{2 \times 10 \times 0.2} [∵ h = 0.2 m]$
$v_{2} = \sqrt{4}$
$v_{2} = 2 m / s$
Put the value $v_{2}$ in equation (i),
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + (4 \times 2)$
$2 \times 6 = (2 \times 10^{- 2}) v + 8$
$(2 \times 10^{- 2}) v = 12 - 8$
$(2 \times 10^{- 2}) v = 4$
$v = \frac{4}{(2 \times 10^{- 2})}$
$v = 2 \times 10^{2}$
$v = 200 m / \sec$

AIIMS-27.05.2018(M)]

Work, Energy and Power

148927 The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 kg$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

1

10 J

2

20 J

3

40 J

4

50 J

Explanation:

A Given,
$\begin{array}{r} Mass of the body (m) = 25 kg \\ Velocity (v) = 2 m / s \end{array}$
Initial kinetic energy (K.E. $)_{1} = \frac{1}{2} {mv}^{2}$
$(K.E.)_{1} = \frac{1}{2} \times 25 \times 2 \times 2$
$(K.E.)_{1} = 50 J$
When the distance covered by the body is $4 m$ , its kinetic energy will be reduced due to resistive force because body will work against this resistive force.
Therefore,
Work done against resistive force
$=$ Area under $F - x$ graph $=$ Area of triangle
$= \frac{1}{2} \times base \times height$
$= \frac{1}{2} \times 4 \times 20 = 40 J$
Final kinetic energy
$=$ initial kinetic energy - work done against resistive force
$= 50 - 40$
$= 10 J$

AP EAMCET-03.09.2021 Shift-I]

Work, Energy and Power

148928 If potential energy is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ . Then find out maximum force. (Given $a = 2, b = 4$ )

1

- \frac{16}{27} N

2

- \frac{32}{27} N

3

+ \frac{32}{27} N

4

+ \frac{16}{27} N

Explanation:

A Given,
The potential energy,
$U = \frac{a}{r^{2}} - \frac{b}{r}$
And, $a = 2, b = 4$
We know that,
Force on the system will be given by
$F = - \frac{d U}{d r}$
$F = - \frac{d}{d r} [\frac{a}{r^{2}} - \frac{b}{r}]$
$F = - \frac{d}{d r} [a^{- 2} - b r^{- 1}]$
$F = - a \frac{d}{d r} r^{- 2} + b \frac{d}{d r} r^{- 1}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Force will be maximum if, $\frac{dF}{dr} = 0$
$\frac{dF}{dr} = 0$
$\Rightarrow \frac{d}{dr} [\frac{2 a}{r^{3}} - \frac{b}{r^{2}}] = 0$
$\Rightarrow 2 a \frac{d}{dr} r^{- 3} - b \frac{d}{dr} r^{- 2} = 0$
$\Rightarrow - 6 {ar}^{- 4} + 2 {br}^{- 3} = 0$
$\Rightarrow - \frac{6 a}{r^{4}} + \frac{2 b}{r^{3}} = 0$
$\Rightarrow \frac{3 a}{r} = 3 a$
$\Rightarrow r = 3 (\frac{a}{b})$
Put the value $(a = 2)$ and $(b = 4)$ in above equation
$\Rightarrow r = 3 (\frac{2}{4})$
$\Rightarrow r = \frac{3}{2}$
Taking equation (i)
$F_{max} = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Put the value $a, b$ and $r$ in above equation.
$F_{max} = \frac{2 \times 2}{{(\frac{3}{2})}^{3}} - \frac{4}{{(\frac{3}{2})}^{2}}$
$F_{max} = \frac{4}{\frac{27}{8}} - \frac{4}{\frac{9}{4}}$
$F_{max} = \frac{4 \times 8}{27} - \frac{4 \times 4}{9}$
$F_{max} = \frac{16}{9} (\frac{2}{3} - 1)$
$F_{max} = - \frac{16}{9 \times 3}$
$F_{max} = - \frac{16}{27} N$

TS EAMCET-04.08.2021 Shift-I]

Work, Energy and Power

148929 A ball of mass $2 kg$ is thrown from a tall building with velocity,
$v = (20 m / s) \hat{i} + (24 m / s) \hat{j}$ at time $t = 0 s$ .
Change in the potential energy of the ball after, $t = 8 s$ is (The ball is assumed to be in air during its motion between $0 s$ and $8 s$ , $\hat{i}$ is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g = 10 m / s^{2}$ )

1

- 2.56 kJ

2

0.52 kJ

3

1.76 kJ

4

- 2.44 kJ

Explanation:

A Given that,
Initial velocity of ball $(u) = (20 \hat{i} + 24 \hat{j}) m / s$ at time $t = 0$ $\sec$
$Mass (m) = 2 kg$
Let $v =$ velocity at time $t = 8 \sec$
Then after time $t = 8 \sec$
Let the equation of motion be,
$v = u - gt$
$v = (20 \hat{i} + 24 \hat{j}) - (10 \hat{j}) \times 8$
$(∵ g always work in vertical direction)$
$v = (20 \hat{i} - 56 \hat{j}) m / s$
From the law of conservation of energy,
Change in potential energy $=$ Change in kinetic energy
$Δ P.E. = Δ K.E.$
$Δ$ P.E. $= \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$
$Δ P.E. = \frac{1}{2} m [u . u - v . v] = \frac{1}{2} \times 2 [u . u - v . v]$
$Δ$ P.E. $= [(20 \hat{i} + 24 \hat{j}) \cdot (20 \hat{i} + 24 \hat{j})] - [(20 \hat{i} - 56 \hat{j}) \cdot (20 \hat{i} - 56 \hat{j})]$
$Δ P.E. = [(20)^{2} + (24)^{2}] - [(20)^{2} + (56)^{2}]$
$Δ$ P.E. $= 576 - 3136$
$Δ$ P.E. $= - 2560 J = - 2.56 kJ$

TS- EAMCET-10.09.2020

Work, Energy and Power

148925 If the linear momentum is increased by $50 %$ , the kinetic energy will increase by:

1

50 %

2

100 %

3

125 %

4

10 %

Explanation:

C Given, linear momentum increased $= 50 %$ .
We know that,
Initial kinetic energy (K.E. $)_{i} = \frac{p_{i}^{2}}{2 m}$
Final kinetic energy $(K.E.)_{f} = \frac{{(1.5 p_{i})}^{2}}{2 m}$
or $(K.E.)_{f} = \frac{2.25 P_{i}^{2}}{2 m} = 2.25 (K.E.)_{i}$
Increase kinetic energy $= \frac{(K.E.)_{f} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= \frac{2.25 (K.E.)_{i} - (K.E.)_{i}}{(K.E.)_{i}} \times 100$
$= 125 %$
Hence, change in K.E. $= 125 %$ of initial

AIIMS-2013

Work, Energy and Power

148926 A bullet of mass $20 g$ and moving with $600 m / s$ collides with a block of mass $4 kg$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 $m$ after collision?

1

200 m / s

2

150 m / s

3

400 m / s

4

300 m / s

Explanation:

A Given,
Mass of bullet $(m_{1}) = 20 g = 20 \times 10^{- 3} kg$
Velocity of bullet before collision $(v_{1}) = 600 m / \sec$
Mass of block $(m_{2}) = 4 kg$
$h = 0.2 m$
According to law of conservation of linear momentum, $m_{1} v_{1} = m_{1} v + m_{2} v_{2}$
$v =$ velocity of bullet after collision
$v_{2} =$ velocity of block
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + 4 v_{2}$
After collision, block rises to height $0.2 m$
From the law of conservation of energy,
$m_{2} gh = \frac{1}{2} m_{2} v_{2}^{2}$
$2 gh = v_{2}^{2}$
$v_{2} = \sqrt{2 gh}$
$v_{2} = \sqrt{2 \times 10 \times 0.2} [∵ h = 0.2 m]$
$v_{2} = \sqrt{4}$
$v_{2} = 2 m / s$
Put the value $v_{2}$ in equation (i),
$(20 \times 10^{- 3}) \times 600 = (20 \times 10^{- 3}) v + (4 \times 2)$
$2 \times 6 = (2 \times 10^{- 2}) v + 8$
$(2 \times 10^{- 2}) v = 12 - 8$
$(2 \times 10^{- 2}) v = 4$
$v = \frac{4}{(2 \times 10^{- 2})}$
$v = 2 \times 10^{2}$
$v = 200 m / \sec$

AIIMS-27.05.2018(M)]

Work, Energy and Power

148927 The graph between the resistive force $F$ acting on a body and the distance covered by the body is shown in the figure. The mass of the body is $25 kg$ and initial velocity is $2 m / s$ . When the distance covered by the body is $4 m$ , its kinetic energy would be

1

10 J

2

20 J

3

40 J

4

50 J

Explanation:

A Given,
$\begin{array}{r} Mass of the body (m) = 25 kg \\ Velocity (v) = 2 m / s \end{array}$
Initial kinetic energy (K.E. $)_{1} = \frac{1}{2} {mv}^{2}$
$(K.E.)_{1} = \frac{1}{2} \times 25 \times 2 \times 2$
$(K.E.)_{1} = 50 J$
When the distance covered by the body is $4 m$ , its kinetic energy will be reduced due to resistive force because body will work against this resistive force.
Therefore,
Work done against resistive force
$=$ Area under $F - x$ graph $=$ Area of triangle
$= \frac{1}{2} \times base \times height$
$= \frac{1}{2} \times 4 \times 20 = 40 J$
Final kinetic energy
$=$ initial kinetic energy - work done against resistive force
$= 50 - 40$
$= 10 J$

AP EAMCET-03.09.2021 Shift-I]

Work, Energy and Power

148928 If potential energy is given by $U = \frac{a}{r^{2}} - \frac{b}{r}$ . Then find out maximum force. (Given $a = 2, b = 4$ )

1

- \frac{16}{27} N

2

- \frac{32}{27} N

3

+ \frac{32}{27} N

4

+ \frac{16}{27} N

Explanation:

A Given,
The potential energy,
$U = \frac{a}{r^{2}} - \frac{b}{r}$
And, $a = 2, b = 4$
We know that,
Force on the system will be given by
$F = - \frac{d U}{d r}$
$F = - \frac{d}{d r} [\frac{a}{r^{2}} - \frac{b}{r}]$
$F = - \frac{d}{d r} [a^{- 2} - b r^{- 1}]$
$F = - a \frac{d}{d r} r^{- 2} + b \frac{d}{d r} r^{- 1}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Force will be maximum if, $\frac{dF}{dr} = 0$
$\frac{dF}{dr} = 0$
$\Rightarrow \frac{d}{dr} [\frac{2 a}{r^{3}} - \frac{b}{r^{2}}] = 0$
$\Rightarrow 2 a \frac{d}{dr} r^{- 3} - b \frac{d}{dr} r^{- 2} = 0$
$\Rightarrow - 6 {ar}^{- 4} + 2 {br}^{- 3} = 0$
$\Rightarrow - \frac{6 a}{r^{4}} + \frac{2 b}{r^{3}} = 0$
$\Rightarrow \frac{3 a}{r} = 3 a$
$\Rightarrow r = 3 (\frac{a}{b})$
Put the value $(a = 2)$ and $(b = 4)$ in above equation
$\Rightarrow r = 3 (\frac{2}{4})$
$\Rightarrow r = \frac{3}{2}$
Taking equation (i)
$F_{max} = \frac{2 a}{r^{3}} - \frac{b}{r^{2}}$
Put the value $a, b$ and $r$ in above equation.
$F_{max} = \frac{2 \times 2}{{(\frac{3}{2})}^{3}} - \frac{4}{{(\frac{3}{2})}^{2}}$
$F_{max} = \frac{4}{\frac{27}{8}} - \frac{4}{\frac{9}{4}}$
$F_{max} = \frac{4 \times 8}{27} - \frac{4 \times 4}{9}$
$F_{max} = \frac{16}{9} (\frac{2}{3} - 1)$
$F_{max} = - \frac{16}{9 \times 3}$
$F_{max} = - \frac{16}{27} N$

TS EAMCET-04.08.2021 Shift-I]

Work, Energy and Power

148929 A ball of mass $2 kg$ is thrown from a tall building with velocity,
$v = (20 m / s) \hat{i} + (24 m / s) \hat{j}$ at time $t = 0 s$ .
Change in the potential energy of the ball after, $t = 8 s$ is (The ball is assumed to be in air during its motion between $0 s$ and $8 s$ , $\hat{i}$ is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g = 10 m / s^{2}$ )

1

- 2.56 kJ

2

0.52 kJ

3

1.76 kJ

4

- 2.44 kJ

Explanation:

A Given that,
Initial velocity of ball $(u) = (20 \hat{i} + 24 \hat{j}) m / s$ at time $t = 0$ $\sec$
$Mass (m) = 2 kg$
Let $v =$ velocity at time $t = 8 \sec$
Then after time $t = 8 \sec$
Let the equation of motion be,
$v = u - gt$
$v = (20 \hat{i} + 24 \hat{j}) - (10 \hat{j}) \times 8$
$(∵ g always work in vertical direction)$
$v = (20 \hat{i} - 56 \hat{j}) m / s$
From the law of conservation of energy,
Change in potential energy $=$ Change in kinetic energy
$Δ P.E. = Δ K.E.$
$Δ$ P.E. $= \frac{1}{2} m u^{2} - \frac{1}{2} m v^{2}$
$Δ P.E. = \frac{1}{2} m [u . u - v . v] = \frac{1}{2} \times 2 [u . u - v . v]$
$Δ$ P.E. $= [(20 \hat{i} + 24 \hat{j}) \cdot (20 \hat{i} + 24 \hat{j})] - [(20 \hat{i} - 56 \hat{j}) \cdot (20 \hat{i} - 56 \hat{j})]$
$Δ P.E. = [(20)^{2} + (24)^{2}] - [(20)^{2} + (56)^{2}]$
$Δ$ P.E. $= 576 - 3136$
$Δ$ P.E. $= - 2560 J = - 2.56 kJ$

TS- EAMCET-10.09.2020