QBManager

Work, Energy and Power

148974 A solid sphere of mass $1 kg$ and radius $10 cm$ rolls without slipping on a horizontal surface, with velocity of $10 cm / s$ . The total kinetic energy of sphere is

1

0.007 J

2

0.05 J

3

0.01 J

4

0.07 J

Explanation:

A Given,
$Mass (m) = 1 kg$
$Radius (r) = 10 cm = 0.1 m .$
$Velocity (v) = 10 cm / s = 0.1 m / s$
The momentum of inertia of solid sphere (I) $= \frac{2}{5} {mr}^{2}$ We know that,
$ω = \frac{v}{r}$
Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy
$K.E. = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{2} \times (\frac{2}{5} {mr}^{2}) \times {(\frac{v}{r})}^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2}$
$= \frac{7}{10} {mv}^{2}$
$= \frac{7}{10} \times 1 \times (0.1)^{2}$
$= 7 \times 10^{- 3}$
$= 0.007 J$
Thus, total kinetic energy is $0.007 J$ .

MHT-CET 2020

Work, Energy and Power

148975 A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be

1

{[\frac{10 gh}{7}]}^{\frac{1}{2}}

2

\sqrt{2 g h}

3

\frac{10 gh}{7}

4

\sqrt{gh}

Explanation:

A We know,
Potential energy $=$ Translational energy + Rotational energy
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
Moment of inertia of solid sphere -
$I = \frac{2}{5} {mr}^{2}$
And\omega=\frac{\mathrm{v}}{\mathrm{r}} $P u t t i n g t h e v a l u e o f$ I $a n d$ \omega $i n e q u a t i o n (i) -$ \therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2} $H e n c e, T o c l i m b t h e i n c l i n e d s u r f a c e v e l o c i t y s h o u l d b e g r e a t e r t h a n,$ \sqrt{\frac{10}{7} \mathrm{gh}}$.

MHT-CET 2020

Work, Energy and Power

148976 A machine which is $70 %$ efficient raised a 10 kg body through a certain distance and spends $100 J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

1 0

2

70 J

3

50 J

4

35 J

Explanation:

B Given, mass (m) $= 10 kg$
Efficiency of machine $= 70 %$
Actual work done by machine,
$W = 70 % \times 100$
$W = 70 Joule$
Now, $W = mgh$
$70 = 10 \times 10 \times h [∵ g = 10 m / s^{2}]$
$\frac{70}{100} = h$
$h = 0.70 m$
Then, applying conservation of energy-
$K . E = P . E$
$K . E = P . E = mgh$
$= 10 \times 10 \times 0.70$
$K . E = 70$ Joule

AP EAMCET (18.09.2020) Shift-II

Work, Energy and Power

148977 A particle is moving on $X$ -axis has potential energy $U = 2 - 20 x + 5 x^{2} J$ along $X$ -axis. The particle is released at $x = - 3$ . The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)

1

5 m

2

3 m

3

7 m

4

8 m

Explanation:

C Given that, $U = 2 - 20 x + 5 x^{2} J$
We know that, Force $(F) = - \frac{dU}{dx}$
$F = - [- 20 + 10 x]$
$F = 20 - 10 x$
If particle is in equilibrium position then $F = 0$
Putting the value of $F$ in equation (i), we get-
$0 = 20 - 10 x$
$20 = 10 x$
$x = 2 m$
The particle is released from $x = - 3$ . So, amplitude from $x = - 3$ and $x = 2$ is $5 m$ .
The maximum value of $x$ will be-
$x = 5 + 2$
$x = 7 m$

AP EAMCET (18.09.2020) Shift-I

Work, Energy and Power

148978 A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to

1

\frac{1}{x}

2

x

3

x^{2}

4

e^{x}

Explanation:

C Given that, particles move in a straight line with its retardation proportional to its displacement.
$a \propto x$
$a = - kx$
Where, $k =$ proportionality constant $vdv = kx dx$
Integrating both side, we get-
$\int_{v_{i}}^{v_{f}} v d v = - \int_{0}^{x} k x d x$
Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively.
${(\frac{v^{2}}{2})}_{v_{i}}^{v_{f}} = - k {(\frac{x^{2}}{2})}_{0}^{x}$
$\frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} = - k \frac{x^{2}}{2}$
$\frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} = - \frac{{mkx}^{2}}{2}$
$(K.E.)_{final} - (K.E.)_{initial} = - \frac{1}{2} {mkx}^{2}$
Hence, loss in kinetic energy $\propto x^{2}$

UPSEE 2019

Work, Energy and Power

148974 A solid sphere of mass $1 kg$ and radius $10 cm$ rolls without slipping on a horizontal surface, with velocity of $10 cm / s$ . The total kinetic energy of sphere is

1

0.007 J

2

0.05 J

3

0.01 J

4

0.07 J

Explanation:

A Given,
$Mass (m) = 1 kg$
$Radius (r) = 10 cm = 0.1 m .$
$Velocity (v) = 10 cm / s = 0.1 m / s$
The momentum of inertia of solid sphere (I) $= \frac{2}{5} {mr}^{2}$ We know that,
$ω = \frac{v}{r}$
Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy
$K.E. = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{2} \times (\frac{2}{5} {mr}^{2}) \times {(\frac{v}{r})}^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2}$
$= \frac{7}{10} {mv}^{2}$
$= \frac{7}{10} \times 1 \times (0.1)^{2}$
$= 7 \times 10^{- 3}$
$= 0.007 J$
Thus, total kinetic energy is $0.007 J$ .

MHT-CET 2020

Work, Energy and Power

148975 A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be

1

{[\frac{10 gh}{7}]}^{\frac{1}{2}}

2

\sqrt{2 g h}

3

\frac{10 gh}{7}

4

\sqrt{gh}

Explanation:

A We know,
Potential energy $=$ Translational energy + Rotational energy
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
Moment of inertia of solid sphere -
$I = \frac{2}{5} {mr}^{2}$
And\omega=\frac{\mathrm{v}}{\mathrm{r}} $P u t t i n g t h e v a l u e o f$ I $a n d$ \omega $i n e q u a t i o n (i) -$ \therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2} $H e n c e, T o c l i m b t h e i n c l i n e d s u r f a c e v e l o c i t y s h o u l d b e g r e a t e r t h a n,$ \sqrt{\frac{10}{7} \mathrm{gh}}$.

MHT-CET 2020

Work, Energy and Power

148976 A machine which is $70 %$ efficient raised a 10 kg body through a certain distance and spends $100 J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

1 0

2

70 J

3

50 J

4

35 J

Explanation:

B Given, mass (m) $= 10 kg$
Efficiency of machine $= 70 %$
Actual work done by machine,
$W = 70 % \times 100$
$W = 70 Joule$
Now, $W = mgh$
$70 = 10 \times 10 \times h [∵ g = 10 m / s^{2}]$
$\frac{70}{100} = h$
$h = 0.70 m$
Then, applying conservation of energy-
$K . E = P . E$
$K . E = P . E = mgh$
$= 10 \times 10 \times 0.70$
$K . E = 70$ Joule

AP EAMCET (18.09.2020) Shift-II

Work, Energy and Power

148977 A particle is moving on $X$ -axis has potential energy $U = 2 - 20 x + 5 x^{2} J$ along $X$ -axis. The particle is released at $x = - 3$ . The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)

1

5 m

2

3 m

3

7 m

4

8 m

Explanation:

C Given that, $U = 2 - 20 x + 5 x^{2} J$
We know that, Force $(F) = - \frac{dU}{dx}$
$F = - [- 20 + 10 x]$
$F = 20 - 10 x$
If particle is in equilibrium position then $F = 0$
Putting the value of $F$ in equation (i), we get-
$0 = 20 - 10 x$
$20 = 10 x$
$x = 2 m$
The particle is released from $x = - 3$ . So, amplitude from $x = - 3$ and $x = 2$ is $5 m$ .
The maximum value of $x$ will be-
$x = 5 + 2$
$x = 7 m$

AP EAMCET (18.09.2020) Shift-I

Work, Energy and Power

148978 A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to

1

\frac{1}{x}

2

x

3

x^{2}

4

e^{x}

Explanation:

C Given that, particles move in a straight line with its retardation proportional to its displacement.
$a \propto x$
$a = - kx$
Where, $k =$ proportionality constant $vdv = kx dx$
Integrating both side, we get-
$\int_{v_{i}}^{v_{f}} v d v = - \int_{0}^{x} k x d x$
Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively.
${(\frac{v^{2}}{2})}_{v_{i}}^{v_{f}} = - k {(\frac{x^{2}}{2})}_{0}^{x}$
$\frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} = - k \frac{x^{2}}{2}$
$\frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} = - \frac{{mkx}^{2}}{2}$
$(K.E.)_{final} - (K.E.)_{initial} = - \frac{1}{2} {mkx}^{2}$
Hence, loss in kinetic energy $\propto x^{2}$

UPSEE 2019

Work, Energy and Power

148974 A solid sphere of mass $1 kg$ and radius $10 cm$ rolls without slipping on a horizontal surface, with velocity of $10 cm / s$ . The total kinetic energy of sphere is

1

0.007 J

2

0.05 J

3

0.01 J

4

0.07 J

Explanation:

A Given,
$Mass (m) = 1 kg$
$Radius (r) = 10 cm = 0.1 m .$
$Velocity (v) = 10 cm / s = 0.1 m / s$
The momentum of inertia of solid sphere (I) $= \frac{2}{5} {mr}^{2}$ We know that,
$ω = \frac{v}{r}$
Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy
$K.E. = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{2} \times (\frac{2}{5} {mr}^{2}) \times {(\frac{v}{r})}^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2}$
$= \frac{7}{10} {mv}^{2}$
$= \frac{7}{10} \times 1 \times (0.1)^{2}$
$= 7 \times 10^{- 3}$
$= 0.007 J$
Thus, total kinetic energy is $0.007 J$ .

MHT-CET 2020

Work, Energy and Power

148975 A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be

1

{[\frac{10 gh}{7}]}^{\frac{1}{2}}

2

\sqrt{2 g h}

3

\frac{10 gh}{7}

4

\sqrt{gh}

Explanation:

A We know,
Potential energy $=$ Translational energy + Rotational energy
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
Moment of inertia of solid sphere -
$I = \frac{2}{5} {mr}^{2}$
And\omega=\frac{\mathrm{v}}{\mathrm{r}} $P u t t i n g t h e v a l u e o f$ I $a n d$ \omega $i n e q u a t i o n (i) -$ \therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2} $H e n c e, T o c l i m b t h e i n c l i n e d s u r f a c e v e l o c i t y s h o u l d b e g r e a t e r t h a n,$ \sqrt{\frac{10}{7} \mathrm{gh}}$.

MHT-CET 2020

Work, Energy and Power

148976 A machine which is $70 %$ efficient raised a 10 kg body through a certain distance and spends $100 J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

1 0

2

70 J

3

50 J

4

35 J

Explanation:

B Given, mass (m) $= 10 kg$
Efficiency of machine $= 70 %$
Actual work done by machine,
$W = 70 % \times 100$
$W = 70 Joule$
Now, $W = mgh$
$70 = 10 \times 10 \times h [∵ g = 10 m / s^{2}]$
$\frac{70}{100} = h$
$h = 0.70 m$
Then, applying conservation of energy-
$K . E = P . E$
$K . E = P . E = mgh$
$= 10 \times 10 \times 0.70$
$K . E = 70$ Joule

AP EAMCET (18.09.2020) Shift-II

Work, Energy and Power

148977 A particle is moving on $X$ -axis has potential energy $U = 2 - 20 x + 5 x^{2} J$ along $X$ -axis. The particle is released at $x = - 3$ . The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)

1

5 m

2

3 m

3

7 m

4

8 m

Explanation:

C Given that, $U = 2 - 20 x + 5 x^{2} J$
We know that, Force $(F) = - \frac{dU}{dx}$
$F = - [- 20 + 10 x]$
$F = 20 - 10 x$
If particle is in equilibrium position then $F = 0$
Putting the value of $F$ in equation (i), we get-
$0 = 20 - 10 x$
$20 = 10 x$
$x = 2 m$
The particle is released from $x = - 3$ . So, amplitude from $x = - 3$ and $x = 2$ is $5 m$ .
The maximum value of $x$ will be-
$x = 5 + 2$
$x = 7 m$

AP EAMCET (18.09.2020) Shift-I

Work, Energy and Power

148978 A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to

1

\frac{1}{x}

2

x

3

x^{2}

4

e^{x}

Explanation:

C Given that, particles move in a straight line with its retardation proportional to its displacement.
$a \propto x$
$a = - kx$
Where, $k =$ proportionality constant $vdv = kx dx$
Integrating both side, we get-
$\int_{v_{i}}^{v_{f}} v d v = - \int_{0}^{x} k x d x$
Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively.
${(\frac{v^{2}}{2})}_{v_{i}}^{v_{f}} = - k {(\frac{x^{2}}{2})}_{0}^{x}$
$\frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} = - k \frac{x^{2}}{2}$
$\frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} = - \frac{{mkx}^{2}}{2}$
$(K.E.)_{final} - (K.E.)_{initial} = - \frac{1}{2} {mkx}^{2}$
Hence, loss in kinetic energy $\propto x^{2}$

UPSEE 2019

Work, Energy and Power

148974 A solid sphere of mass $1 kg$ and radius $10 cm$ rolls without slipping on a horizontal surface, with velocity of $10 cm / s$ . The total kinetic energy of sphere is

1

0.007 J

2

0.05 J

3

0.01 J

4

0.07 J

Explanation:

A Given,
$Mass (m) = 1 kg$
$Radius (r) = 10 cm = 0.1 m .$
$Velocity (v) = 10 cm / s = 0.1 m / s$
The momentum of inertia of solid sphere (I) $= \frac{2}{5} {mr}^{2}$ We know that,
$ω = \frac{v}{r}$
Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy
$K.E. = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{2} \times (\frac{2}{5} {mr}^{2}) \times {(\frac{v}{r})}^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2}$
$= \frac{7}{10} {mv}^{2}$
$= \frac{7}{10} \times 1 \times (0.1)^{2}$
$= 7 \times 10^{- 3}$
$= 0.007 J$
Thus, total kinetic energy is $0.007 J$ .

MHT-CET 2020

Work, Energy and Power

148975 A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be

1

{[\frac{10 gh}{7}]}^{\frac{1}{2}}

2

\sqrt{2 g h}

3

\frac{10 gh}{7}

4

\sqrt{gh}

Explanation:

A We know,
Potential energy $=$ Translational energy + Rotational energy
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
Moment of inertia of solid sphere -
$I = \frac{2}{5} {mr}^{2}$
And\omega=\frac{\mathrm{v}}{\mathrm{r}} $P u t t i n g t h e v a l u e o f$ I $a n d$ \omega $i n e q u a t i o n (i) -$ \therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2} $H e n c e, T o c l i m b t h e i n c l i n e d s u r f a c e v e l o c i t y s h o u l d b e g r e a t e r t h a n,$ \sqrt{\frac{10}{7} \mathrm{gh}}$.

MHT-CET 2020

Work, Energy and Power

148976 A machine which is $70 %$ efficient raised a 10 kg body through a certain distance and spends $100 J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

1 0

2

70 J

3

50 J

4

35 J

Explanation:

B Given, mass (m) $= 10 kg$
Efficiency of machine $= 70 %$
Actual work done by machine,
$W = 70 % \times 100$
$W = 70 Joule$
Now, $W = mgh$
$70 = 10 \times 10 \times h [∵ g = 10 m / s^{2}]$
$\frac{70}{100} = h$
$h = 0.70 m$
Then, applying conservation of energy-
$K . E = P . E$
$K . E = P . E = mgh$
$= 10 \times 10 \times 0.70$
$K . E = 70$ Joule

AP EAMCET (18.09.2020) Shift-II

Work, Energy and Power

148977 A particle is moving on $X$ -axis has potential energy $U = 2 - 20 x + 5 x^{2} J$ along $X$ -axis. The particle is released at $x = - 3$ . The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)

1

5 m

2

3 m

3

7 m

4

8 m

Explanation:

C Given that, $U = 2 - 20 x + 5 x^{2} J$
We know that, Force $(F) = - \frac{dU}{dx}$
$F = - [- 20 + 10 x]$
$F = 20 - 10 x$
If particle is in equilibrium position then $F = 0$
Putting the value of $F$ in equation (i), we get-
$0 = 20 - 10 x$
$20 = 10 x$
$x = 2 m$
The particle is released from $x = - 3$ . So, amplitude from $x = - 3$ and $x = 2$ is $5 m$ .
The maximum value of $x$ will be-
$x = 5 + 2$
$x = 7 m$

AP EAMCET (18.09.2020) Shift-I

Work, Energy and Power

148978 A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to

1

\frac{1}{x}

2

x

3

x^{2}

4

e^{x}

Explanation:

C Given that, particles move in a straight line with its retardation proportional to its displacement.
$a \propto x$
$a = - kx$
Where, $k =$ proportionality constant $vdv = kx dx$
Integrating both side, we get-
$\int_{v_{i}}^{v_{f}} v d v = - \int_{0}^{x} k x d x$
Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively.
${(\frac{v^{2}}{2})}_{v_{i}}^{v_{f}} = - k {(\frac{x^{2}}{2})}_{0}^{x}$
$\frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} = - k \frac{x^{2}}{2}$
$\frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} = - \frac{{mkx}^{2}}{2}$
$(K.E.)_{final} - (K.E.)_{initial} = - \frac{1}{2} {mkx}^{2}$
Hence, loss in kinetic energy $\propto x^{2}$

UPSEE 2019

Work, Energy and Power

148974 A solid sphere of mass $1 kg$ and radius $10 cm$ rolls without slipping on a horizontal surface, with velocity of $10 cm / s$ . The total kinetic energy of sphere is

1

0.007 J

2

0.05 J

3

0.01 J

4

0.07 J

Explanation:

A Given,
$Mass (m) = 1 kg$
$Radius (r) = 10 cm = 0.1 m .$
$Velocity (v) = 10 cm / s = 0.1 m / s$
The momentum of inertia of solid sphere (I) $= \frac{2}{5} {mr}^{2}$ We know that,
$ω = \frac{v}{r}$
Total kinetic energy of a rolling body without slipping (K.E.) $=$ Rotational kinetic energy + linear kinetic energy
$K.E. = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{2} \times (\frac{2}{5} {mr}^{2}) \times {(\frac{v}{r})}^{2}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mr}^{2} \times \frac{v^{2}}{r^{2}}$
$= \frac{1}{2} {mv}^{2} + \frac{1}{5} {mv}^{2}$
$= \frac{7}{10} {mv}^{2}$
$= \frac{7}{10} \times 1 \times (0.1)^{2}$
$= 7 \times 10^{- 3}$
$= 0.007 J$
Thus, total kinetic energy is $0.007 J$ .

MHT-CET 2020

Work, Energy and Power

148975 A solid sphere is rolling on a frictionless surface with translational velocity ' $v$ '. It climbs that inclined plane from ' $A$ ' to ' $B$ ' and then moves away from $B$ on the smooth horizontal surface. The value of ' $v$ ' should be

1

{[\frac{10 gh}{7}]}^{\frac{1}{2}}

2

\sqrt{2 g h}

3

\frac{10 gh}{7}

4

\sqrt{gh}

Explanation:

A We know,
Potential energy $=$ Translational energy + Rotational energy
$\frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2} = mgh$
Moment of inertia of solid sphere -
$I = \frac{2}{5} {mr}^{2}$
And\omega=\frac{\mathrm{v}}{\mathrm{r}} $P u t t i n g t h e v a l u e o f$ I $a n d$ \omega $i n e q u a t i o n (i) -$ \therefore \frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh}\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=m g h\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}\mathrm{v}^{2}\left(\frac{7}{10}\right)=\mathrm{gh}\mathrm{v}^{2}=\frac{10}{7} \mathrm{gh}\mathrm{v}=\sqrt{\frac{10}{7} \mathrm{gh}}=\left[\frac{10 \mathrm{gh}}{7}\right]^{1 / 2} $H e n c e, T o c l i m b t h e i n c l i n e d s u r f a c e v e l o c i t y s h o u l d b e g r e a t e r t h a n,$ \sqrt{\frac{10}{7} \mathrm{gh}}$.

MHT-CET 2020

Work, Energy and Power

148976 A machine which is $70 %$ efficient raised a 10 kg body through a certain distance and spends $100 J$ energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

1 0

2

70 J

3

50 J

4

35 J

Explanation:

B Given, mass (m) $= 10 kg$
Efficiency of machine $= 70 %$
Actual work done by machine,
$W = 70 % \times 100$
$W = 70 Joule$
Now, $W = mgh$
$70 = 10 \times 10 \times h [∵ g = 10 m / s^{2}]$
$\frac{70}{100} = h$
$h = 0.70 m$
Then, applying conservation of energy-
$K . E = P . E$
$K . E = P . E = mgh$
$= 10 \times 10 \times 0.70$
$K . E = 70$ Joule

AP EAMCET (18.09.2020) Shift-II

Work, Energy and Power

148977 A particle is moving on $X$ -axis has potential energy $U = 2 - 20 x + 5 x^{2} J$ along $X$ -axis. The particle is released at $x = - 3$ . The maximum value of $x$ will be ( $x$ is in metre and $U$ is in Joules)

1

5 m

2

3 m

3

7 m

4

8 m

Explanation:

C Given that, $U = 2 - 20 x + 5 x^{2} J$
We know that, Force $(F) = - \frac{dU}{dx}$
$F = - [- 20 + 10 x]$
$F = 20 - 10 x$
If particle is in equilibrium position then $F = 0$
Putting the value of $F$ in equation (i), we get-
$0 = 20 - 10 x$
$20 = 10 x$
$x = 2 m$
The particle is released from $x = - 3$ . So, amplitude from $x = - 3$ and $x = 2$ is $5 m$ .
The maximum value of $x$ will be-
$x = 5 + 2$
$x = 7 m$

AP EAMCET (18.09.2020) Shift-I

Work, Energy and Power

148978 A particle moves in a straight line with its retardation proportional to its displacement. The loss of its kinetic energy for any displacement $x$ is proportional to

1

\frac{1}{x}

2

x

3

x^{2}

4

e^{x}

Explanation:

C Given that, particles move in a straight line with its retardation proportional to its displacement.
$a \propto x$
$a = - kx$
Where, $k =$ proportionality constant $vdv = kx dx$
Integrating both side, we get-
$\int_{v_{i}}^{v_{f}} v d v = - \int_{0}^{x} k x d x$
Where $v_{i}$ and $v_{f}$ are initial and final velocity of particle respectively.
${(\frac{v^{2}}{2})}_{v_{i}}^{v_{f}} = - k {(\frac{x^{2}}{2})}_{0}^{x}$
$\frac{v_{f}^{2}}{2} - \frac{v_{i}^{2}}{2} = - k \frac{x^{2}}{2}$
$\frac{1}{2} {mv}_{f}^{2} - \frac{1}{2} {mv}_{i}^{2} = - \frac{{mkx}^{2}}{2}$
$(K.E.)_{final} - (K.E.)_{initial} = - \frac{1}{2} {mkx}^{2}$
Hence, loss in kinetic energy $\propto x^{2}$

UPSEE 2019