149098
Power applied to a particle varies with time as $\left(P=3 t^{2}-2 t+1\right) W$, where $t$ is in second. The change in its kinetic energy between $t=2 s$ and $\mathrm{t}=\mathbf{4} \mathrm{s}$ is
1 $32 \mathrm{~J}$
2 $46 \mathrm{~J}$
3 $61 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B Given, Power $(\mathrm{P})=3 \mathrm{t}^{2}-2 \mathrm{t}+1$ watt Since, power is time rate of change of work (We know that), $P=\frac{d W}{d t}$ $\mathrm{dW}=\mathrm{P} . \mathrm{dt}$ from work energy theorem Work done $(\mathrm{dW})=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}=\Delta$ K.E. From equation (i) \& (ii), we get - $\Delta \text { K.E. } =P d t$ $\Delta \text { K.E. } =\int_{t=2}^{t=4} P d t$ $=\int_{t=2}^{t=4}\left(3 t^{2}-2 t+1\right) d t$ $=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=2}^{t=4}$ $=\left[t^{3}-t^{2}+t\right]_{t=2}^{t=4}$ $=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]$ $= 64-8-16+4+4-2=46 \mathrm{~J}$
CG PET 2019
Work, Energy and Power
149099
A solid cylinder of mass $m$ and $R$ rolls down plane of height $30 \mathrm{~m}$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $\text { [use } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \text { ] }$
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $30 \mathrm{~m} / \mathrm{s}$
4 $40 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~h}=30 \mathrm{~m}$ Mass of cylinder $=m$ Radius of cylinder $=\mathrm{R}$ At the case of rolling cylinder $\mathrm{v}=\mathrm{R} \omega$ And $\mathrm{I}=\frac{1}{2} \mathrm{mR}^{2}$ To apply work energy theorem, Potential energy $=$ Kinetic energy + Rotational energy $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (ii), we get- $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \omega^{2}$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{~m}(\mathrm{R} \omega)^{2}$ [From equation (i)] $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{mv}^{2}$ $\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{4}{3} \mathrm{gh}}$ $\mathrm{v}=\sqrt{\frac{4}{3} \times 10 \times 30}$ $\mathrm{v}=\sqrt{400}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$
TS EAMCET 18.07.2022
Work, Energy and Power
149100
A ball with $10^{3} \mathrm{~J}$ of kinetic energy collides with a horizontally mounted spring. If the maximum compression of the spring is $50 \mathrm{~cm}$, then the spring constant of the spring is
1 $2 \times 10^{3} \mathrm{Nm}^{-1}$
2 $6 \times 10^{3} \mathrm{Nm}^{-1}$
3 $8 \times 10^{3} \mathrm{Nm}^{-1}$
4 $5 \times 10^{3} \mathrm{Nm}^{-1}$
5 $3 \times 10^{3} \mathrm{Nm}^{-1}$
Explanation:
C Given, Kinetic energy $=10^{3} \mathrm{~J}$ Compression of $\operatorname{spring}\left(\mathrm{x}_{\mathrm{m}}\right)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$ Let spring constant $=\mathrm{K} \mathrm{N} / \mathrm{m}$ At maximum compression $\mathrm{x}_{\mathrm{m}}$, the kinetic energy of the ball is converted entirely into the potential energy of the spring. $\text { K.E. }=\frac{1}{2} \mathrm{Kx}_{\mathrm{m}}^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K}\left(50 \times 10^{-2}\right)^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K} \times\left(2500 \times 10^{-4}\right)$ $\mathrm{K}=\frac{1000 \times 2}{2500 \times 10^{-4}}$ $\mathrm{~K}=8 \times 10^{3} \mathrm{~N} / \mathrm{m}$
Kerala CEE 2021
Work, Energy and Power
149101
An object released from certain height $h$ from the ground rebounds to a height $\frac{h}{4}$ after striking the ground. The fraction of the energy lost by it is
1 $\frac{1}{4}$
2 $\frac{3}{4}$
3 $\frac{1}{2}$
4 $\frac{1}{8}$
5 $\frac{3}{8}$
Explanation:
B - Height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball fall down. $(\text { K.E. })_{\mathrm{i}}=\mathrm{mgh}$ $- When the ball finally hits the ground all the potential energy has convert to kinetic energy.$ $(\text { K.E. })_{\mathrm{f}}=\frac{\mathrm{mgh}}{4}$ Fraction of the energy loss $\Delta \mathrm{K}=\frac{(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}}{(\text { K.E. })_{\mathrm{i}}}$ $\Delta \mathrm{K} =\frac{\mathrm{mgh}-\frac{\mathrm{mgh}}{4}}{\mathrm{mgh}}$ $\Delta \mathrm{K} =\frac{3}{4}$
Kerala CEE 2021
Work, Energy and Power
149102
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, the average resistance offered by the sand to the body is
B Work energy theorem when body reaches sand floor Potential energy $=$ kinetic energy $\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}$ $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ When body penetrates to the sand it's KE will be dissipated by the resultant resistance force $\frac{1}{2} \mathrm{Mv}^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\frac{1}{2} \mathrm{M}(\sqrt{2 \mathrm{gh}})^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\mathrm{Mgh}+\mathrm{Mgx}=\mathrm{Fx}$ $\mathrm{F}=\operatorname{Mg}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)$
149098
Power applied to a particle varies with time as $\left(P=3 t^{2}-2 t+1\right) W$, where $t$ is in second. The change in its kinetic energy between $t=2 s$ and $\mathrm{t}=\mathbf{4} \mathrm{s}$ is
1 $32 \mathrm{~J}$
2 $46 \mathrm{~J}$
3 $61 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B Given, Power $(\mathrm{P})=3 \mathrm{t}^{2}-2 \mathrm{t}+1$ watt Since, power is time rate of change of work (We know that), $P=\frac{d W}{d t}$ $\mathrm{dW}=\mathrm{P} . \mathrm{dt}$ from work energy theorem Work done $(\mathrm{dW})=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}=\Delta$ K.E. From equation (i) \& (ii), we get - $\Delta \text { K.E. } =P d t$ $\Delta \text { K.E. } =\int_{t=2}^{t=4} P d t$ $=\int_{t=2}^{t=4}\left(3 t^{2}-2 t+1\right) d t$ $=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=2}^{t=4}$ $=\left[t^{3}-t^{2}+t\right]_{t=2}^{t=4}$ $=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]$ $= 64-8-16+4+4-2=46 \mathrm{~J}$
CG PET 2019
Work, Energy and Power
149099
A solid cylinder of mass $m$ and $R$ rolls down plane of height $30 \mathrm{~m}$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $\text { [use } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \text { ] }$
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $30 \mathrm{~m} / \mathrm{s}$
4 $40 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~h}=30 \mathrm{~m}$ Mass of cylinder $=m$ Radius of cylinder $=\mathrm{R}$ At the case of rolling cylinder $\mathrm{v}=\mathrm{R} \omega$ And $\mathrm{I}=\frac{1}{2} \mathrm{mR}^{2}$ To apply work energy theorem, Potential energy $=$ Kinetic energy + Rotational energy $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (ii), we get- $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \omega^{2}$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{~m}(\mathrm{R} \omega)^{2}$ [From equation (i)] $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{mv}^{2}$ $\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{4}{3} \mathrm{gh}}$ $\mathrm{v}=\sqrt{\frac{4}{3} \times 10 \times 30}$ $\mathrm{v}=\sqrt{400}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$
TS EAMCET 18.07.2022
Work, Energy and Power
149100
A ball with $10^{3} \mathrm{~J}$ of kinetic energy collides with a horizontally mounted spring. If the maximum compression of the spring is $50 \mathrm{~cm}$, then the spring constant of the spring is
1 $2 \times 10^{3} \mathrm{Nm}^{-1}$
2 $6 \times 10^{3} \mathrm{Nm}^{-1}$
3 $8 \times 10^{3} \mathrm{Nm}^{-1}$
4 $5 \times 10^{3} \mathrm{Nm}^{-1}$
5 $3 \times 10^{3} \mathrm{Nm}^{-1}$
Explanation:
C Given, Kinetic energy $=10^{3} \mathrm{~J}$ Compression of $\operatorname{spring}\left(\mathrm{x}_{\mathrm{m}}\right)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$ Let spring constant $=\mathrm{K} \mathrm{N} / \mathrm{m}$ At maximum compression $\mathrm{x}_{\mathrm{m}}$, the kinetic energy of the ball is converted entirely into the potential energy of the spring. $\text { K.E. }=\frac{1}{2} \mathrm{Kx}_{\mathrm{m}}^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K}\left(50 \times 10^{-2}\right)^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K} \times\left(2500 \times 10^{-4}\right)$ $\mathrm{K}=\frac{1000 \times 2}{2500 \times 10^{-4}}$ $\mathrm{~K}=8 \times 10^{3} \mathrm{~N} / \mathrm{m}$
Kerala CEE 2021
Work, Energy and Power
149101
An object released from certain height $h$ from the ground rebounds to a height $\frac{h}{4}$ after striking the ground. The fraction of the energy lost by it is
1 $\frac{1}{4}$
2 $\frac{3}{4}$
3 $\frac{1}{2}$
4 $\frac{1}{8}$
5 $\frac{3}{8}$
Explanation:
B - Height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball fall down. $(\text { K.E. })_{\mathrm{i}}=\mathrm{mgh}$ $- When the ball finally hits the ground all the potential energy has convert to kinetic energy.$ $(\text { K.E. })_{\mathrm{f}}=\frac{\mathrm{mgh}}{4}$ Fraction of the energy loss $\Delta \mathrm{K}=\frac{(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}}{(\text { K.E. })_{\mathrm{i}}}$ $\Delta \mathrm{K} =\frac{\mathrm{mgh}-\frac{\mathrm{mgh}}{4}}{\mathrm{mgh}}$ $\Delta \mathrm{K} =\frac{3}{4}$
Kerala CEE 2021
Work, Energy and Power
149102
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, the average resistance offered by the sand to the body is
B Work energy theorem when body reaches sand floor Potential energy $=$ kinetic energy $\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}$ $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ When body penetrates to the sand it's KE will be dissipated by the resultant resistance force $\frac{1}{2} \mathrm{Mv}^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\frac{1}{2} \mathrm{M}(\sqrt{2 \mathrm{gh}})^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\mathrm{Mgh}+\mathrm{Mgx}=\mathrm{Fx}$ $\mathrm{F}=\operatorname{Mg}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)$
149098
Power applied to a particle varies with time as $\left(P=3 t^{2}-2 t+1\right) W$, where $t$ is in second. The change in its kinetic energy between $t=2 s$ and $\mathrm{t}=\mathbf{4} \mathrm{s}$ is
1 $32 \mathrm{~J}$
2 $46 \mathrm{~J}$
3 $61 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B Given, Power $(\mathrm{P})=3 \mathrm{t}^{2}-2 \mathrm{t}+1$ watt Since, power is time rate of change of work (We know that), $P=\frac{d W}{d t}$ $\mathrm{dW}=\mathrm{P} . \mathrm{dt}$ from work energy theorem Work done $(\mathrm{dW})=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}=\Delta$ K.E. From equation (i) \& (ii), we get - $\Delta \text { K.E. } =P d t$ $\Delta \text { K.E. } =\int_{t=2}^{t=4} P d t$ $=\int_{t=2}^{t=4}\left(3 t^{2}-2 t+1\right) d t$ $=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=2}^{t=4}$ $=\left[t^{3}-t^{2}+t\right]_{t=2}^{t=4}$ $=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]$ $= 64-8-16+4+4-2=46 \mathrm{~J}$
CG PET 2019
Work, Energy and Power
149099
A solid cylinder of mass $m$ and $R$ rolls down plane of height $30 \mathrm{~m}$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $\text { [use } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \text { ] }$
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $30 \mathrm{~m} / \mathrm{s}$
4 $40 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~h}=30 \mathrm{~m}$ Mass of cylinder $=m$ Radius of cylinder $=\mathrm{R}$ At the case of rolling cylinder $\mathrm{v}=\mathrm{R} \omega$ And $\mathrm{I}=\frac{1}{2} \mathrm{mR}^{2}$ To apply work energy theorem, Potential energy $=$ Kinetic energy + Rotational energy $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (ii), we get- $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \omega^{2}$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{~m}(\mathrm{R} \omega)^{2}$ [From equation (i)] $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{mv}^{2}$ $\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{4}{3} \mathrm{gh}}$ $\mathrm{v}=\sqrt{\frac{4}{3} \times 10 \times 30}$ $\mathrm{v}=\sqrt{400}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$
TS EAMCET 18.07.2022
Work, Energy and Power
149100
A ball with $10^{3} \mathrm{~J}$ of kinetic energy collides with a horizontally mounted spring. If the maximum compression of the spring is $50 \mathrm{~cm}$, then the spring constant of the spring is
1 $2 \times 10^{3} \mathrm{Nm}^{-1}$
2 $6 \times 10^{3} \mathrm{Nm}^{-1}$
3 $8 \times 10^{3} \mathrm{Nm}^{-1}$
4 $5 \times 10^{3} \mathrm{Nm}^{-1}$
5 $3 \times 10^{3} \mathrm{Nm}^{-1}$
Explanation:
C Given, Kinetic energy $=10^{3} \mathrm{~J}$ Compression of $\operatorname{spring}\left(\mathrm{x}_{\mathrm{m}}\right)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$ Let spring constant $=\mathrm{K} \mathrm{N} / \mathrm{m}$ At maximum compression $\mathrm{x}_{\mathrm{m}}$, the kinetic energy of the ball is converted entirely into the potential energy of the spring. $\text { K.E. }=\frac{1}{2} \mathrm{Kx}_{\mathrm{m}}^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K}\left(50 \times 10^{-2}\right)^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K} \times\left(2500 \times 10^{-4}\right)$ $\mathrm{K}=\frac{1000 \times 2}{2500 \times 10^{-4}}$ $\mathrm{~K}=8 \times 10^{3} \mathrm{~N} / \mathrm{m}$
Kerala CEE 2021
Work, Energy and Power
149101
An object released from certain height $h$ from the ground rebounds to a height $\frac{h}{4}$ after striking the ground. The fraction of the energy lost by it is
1 $\frac{1}{4}$
2 $\frac{3}{4}$
3 $\frac{1}{2}$
4 $\frac{1}{8}$
5 $\frac{3}{8}$
Explanation:
B - Height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball fall down. $(\text { K.E. })_{\mathrm{i}}=\mathrm{mgh}$ $- When the ball finally hits the ground all the potential energy has convert to kinetic energy.$ $(\text { K.E. })_{\mathrm{f}}=\frac{\mathrm{mgh}}{4}$ Fraction of the energy loss $\Delta \mathrm{K}=\frac{(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}}{(\text { K.E. })_{\mathrm{i}}}$ $\Delta \mathrm{K} =\frac{\mathrm{mgh}-\frac{\mathrm{mgh}}{4}}{\mathrm{mgh}}$ $\Delta \mathrm{K} =\frac{3}{4}$
Kerala CEE 2021
Work, Energy and Power
149102
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, the average resistance offered by the sand to the body is
B Work energy theorem when body reaches sand floor Potential energy $=$ kinetic energy $\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}$ $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ When body penetrates to the sand it's KE will be dissipated by the resultant resistance force $\frac{1}{2} \mathrm{Mv}^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\frac{1}{2} \mathrm{M}(\sqrt{2 \mathrm{gh}})^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\mathrm{Mgh}+\mathrm{Mgx}=\mathrm{Fx}$ $\mathrm{F}=\operatorname{Mg}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)$
149098
Power applied to a particle varies with time as $\left(P=3 t^{2}-2 t+1\right) W$, where $t$ is in second. The change in its kinetic energy between $t=2 s$ and $\mathrm{t}=\mathbf{4} \mathrm{s}$ is
1 $32 \mathrm{~J}$
2 $46 \mathrm{~J}$
3 $61 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B Given, Power $(\mathrm{P})=3 \mathrm{t}^{2}-2 \mathrm{t}+1$ watt Since, power is time rate of change of work (We know that), $P=\frac{d W}{d t}$ $\mathrm{dW}=\mathrm{P} . \mathrm{dt}$ from work energy theorem Work done $(\mathrm{dW})=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}=\Delta$ K.E. From equation (i) \& (ii), we get - $\Delta \text { K.E. } =P d t$ $\Delta \text { K.E. } =\int_{t=2}^{t=4} P d t$ $=\int_{t=2}^{t=4}\left(3 t^{2}-2 t+1\right) d t$ $=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=2}^{t=4}$ $=\left[t^{3}-t^{2}+t\right]_{t=2}^{t=4}$ $=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]$ $= 64-8-16+4+4-2=46 \mathrm{~J}$
CG PET 2019
Work, Energy and Power
149099
A solid cylinder of mass $m$ and $R$ rolls down plane of height $30 \mathrm{~m}$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $\text { [use } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \text { ] }$
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $30 \mathrm{~m} / \mathrm{s}$
4 $40 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~h}=30 \mathrm{~m}$ Mass of cylinder $=m$ Radius of cylinder $=\mathrm{R}$ At the case of rolling cylinder $\mathrm{v}=\mathrm{R} \omega$ And $\mathrm{I}=\frac{1}{2} \mathrm{mR}^{2}$ To apply work energy theorem, Potential energy $=$ Kinetic energy + Rotational energy $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (ii), we get- $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \omega^{2}$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{~m}(\mathrm{R} \omega)^{2}$ [From equation (i)] $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{mv}^{2}$ $\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{4}{3} \mathrm{gh}}$ $\mathrm{v}=\sqrt{\frac{4}{3} \times 10 \times 30}$ $\mathrm{v}=\sqrt{400}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$
TS EAMCET 18.07.2022
Work, Energy and Power
149100
A ball with $10^{3} \mathrm{~J}$ of kinetic energy collides with a horizontally mounted spring. If the maximum compression of the spring is $50 \mathrm{~cm}$, then the spring constant of the spring is
1 $2 \times 10^{3} \mathrm{Nm}^{-1}$
2 $6 \times 10^{3} \mathrm{Nm}^{-1}$
3 $8 \times 10^{3} \mathrm{Nm}^{-1}$
4 $5 \times 10^{3} \mathrm{Nm}^{-1}$
5 $3 \times 10^{3} \mathrm{Nm}^{-1}$
Explanation:
C Given, Kinetic energy $=10^{3} \mathrm{~J}$ Compression of $\operatorname{spring}\left(\mathrm{x}_{\mathrm{m}}\right)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$ Let spring constant $=\mathrm{K} \mathrm{N} / \mathrm{m}$ At maximum compression $\mathrm{x}_{\mathrm{m}}$, the kinetic energy of the ball is converted entirely into the potential energy of the spring. $\text { K.E. }=\frac{1}{2} \mathrm{Kx}_{\mathrm{m}}^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K}\left(50 \times 10^{-2}\right)^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K} \times\left(2500 \times 10^{-4}\right)$ $\mathrm{K}=\frac{1000 \times 2}{2500 \times 10^{-4}}$ $\mathrm{~K}=8 \times 10^{3} \mathrm{~N} / \mathrm{m}$
Kerala CEE 2021
Work, Energy and Power
149101
An object released from certain height $h$ from the ground rebounds to a height $\frac{h}{4}$ after striking the ground. The fraction of the energy lost by it is
1 $\frac{1}{4}$
2 $\frac{3}{4}$
3 $\frac{1}{2}$
4 $\frac{1}{8}$
5 $\frac{3}{8}$
Explanation:
B - Height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball fall down. $(\text { K.E. })_{\mathrm{i}}=\mathrm{mgh}$ $- When the ball finally hits the ground all the potential energy has convert to kinetic energy.$ $(\text { K.E. })_{\mathrm{f}}=\frac{\mathrm{mgh}}{4}$ Fraction of the energy loss $\Delta \mathrm{K}=\frac{(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}}{(\text { K.E. })_{\mathrm{i}}}$ $\Delta \mathrm{K} =\frac{\mathrm{mgh}-\frac{\mathrm{mgh}}{4}}{\mathrm{mgh}}$ $\Delta \mathrm{K} =\frac{3}{4}$
Kerala CEE 2021
Work, Energy and Power
149102
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, the average resistance offered by the sand to the body is
B Work energy theorem when body reaches sand floor Potential energy $=$ kinetic energy $\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}$ $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ When body penetrates to the sand it's KE will be dissipated by the resultant resistance force $\frac{1}{2} \mathrm{Mv}^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\frac{1}{2} \mathrm{M}(\sqrt{2 \mathrm{gh}})^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\mathrm{Mgh}+\mathrm{Mgx}=\mathrm{Fx}$ $\mathrm{F}=\operatorname{Mg}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)$
149098
Power applied to a particle varies with time as $\left(P=3 t^{2}-2 t+1\right) W$, where $t$ is in second. The change in its kinetic energy between $t=2 s$ and $\mathrm{t}=\mathbf{4} \mathrm{s}$ is
1 $32 \mathrm{~J}$
2 $46 \mathrm{~J}$
3 $61 \mathrm{~J}$
4 $100 \mathrm{~J}$
Explanation:
B Given, Power $(\mathrm{P})=3 \mathrm{t}^{2}-2 \mathrm{t}+1$ watt Since, power is time rate of change of work (We know that), $P=\frac{d W}{d t}$ $\mathrm{dW}=\mathrm{P} . \mathrm{dt}$ from work energy theorem Work done $(\mathrm{dW})=(\text { K.E. })_{\mathrm{f}}-(\text { K.E. })_{\mathrm{i}}=\Delta$ K.E. From equation (i) \& (ii), we get - $\Delta \text { K.E. } =P d t$ $\Delta \text { K.E. } =\int_{t=2}^{t=4} P d t$ $=\int_{t=2}^{t=4}\left(3 t^{2}-2 t+1\right) d t$ $=\left[\frac{3 t^{3}}{3}-\frac{2 t^{2}}{2}+t\right]_{t=2}^{t=4}$ $=\left[t^{3}-t^{2}+t\right]_{t=2}^{t=4}$ $=\left[\left(4^{3}-2^{3}\right)-\left(4^{2}-2^{2}\right)+(4-2)\right]$ $= 64-8-16+4+4-2=46 \mathrm{~J}$
CG PET 2019
Work, Energy and Power
149099
A solid cylinder of mass $m$ and $R$ rolls down plane of height $30 \mathrm{~m}$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $\text { [use } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \text { ] }$
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $30 \mathrm{~m} / \mathrm{s}$
4 $40 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~h}=30 \mathrm{~m}$ Mass of cylinder $=m$ Radius of cylinder $=\mathrm{R}$ At the case of rolling cylinder $\mathrm{v}=\mathrm{R} \omega$ And $\mathrm{I}=\frac{1}{2} \mathrm{mR}^{2}$ To apply work energy theorem, Potential energy $=$ Kinetic energy + Rotational energy $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (ii), we get- $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{1}{2} \mathrm{mR}^{2} \omega^{2}$ $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{~m}(\mathrm{R} \omega)^{2}$ [From equation (i)] $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{4} \mathrm{mv}^{2}$ $\mathrm{mgh}=\frac{3}{4} \mathrm{mv}^{2}$ $\mathrm{v}=\sqrt{\frac{4}{3} \mathrm{gh}}$ $\mathrm{v}=\sqrt{\frac{4}{3} \times 10 \times 30}$ $\mathrm{v}=\sqrt{400}$ $\mathrm{v}=20 \mathrm{~m} / \mathrm{s}$
TS EAMCET 18.07.2022
Work, Energy and Power
149100
A ball with $10^{3} \mathrm{~J}$ of kinetic energy collides with a horizontally mounted spring. If the maximum compression of the spring is $50 \mathrm{~cm}$, then the spring constant of the spring is
1 $2 \times 10^{3} \mathrm{Nm}^{-1}$
2 $6 \times 10^{3} \mathrm{Nm}^{-1}$
3 $8 \times 10^{3} \mathrm{Nm}^{-1}$
4 $5 \times 10^{3} \mathrm{Nm}^{-1}$
5 $3 \times 10^{3} \mathrm{Nm}^{-1}$
Explanation:
C Given, Kinetic energy $=10^{3} \mathrm{~J}$ Compression of $\operatorname{spring}\left(\mathrm{x}_{\mathrm{m}}\right)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$ Let spring constant $=\mathrm{K} \mathrm{N} / \mathrm{m}$ At maximum compression $\mathrm{x}_{\mathrm{m}}$, the kinetic energy of the ball is converted entirely into the potential energy of the spring. $\text { K.E. }=\frac{1}{2} \mathrm{Kx}_{\mathrm{m}}^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K}\left(50 \times 10^{-2}\right)^{2}$ $10^{3}=\frac{1}{2} \mathrm{~K} \times\left(2500 \times 10^{-4}\right)$ $\mathrm{K}=\frac{1000 \times 2}{2500 \times 10^{-4}}$ $\mathrm{~K}=8 \times 10^{3} \mathrm{~N} / \mathrm{m}$
Kerala CEE 2021
Work, Energy and Power
149101
An object released from certain height $h$ from the ground rebounds to a height $\frac{h}{4}$ after striking the ground. The fraction of the energy lost by it is
1 $\frac{1}{4}$
2 $\frac{3}{4}$
3 $\frac{1}{2}$
4 $\frac{1}{8}$
5 $\frac{3}{8}$
Explanation:
B - Height from which the ball falls gives it a potential energy which converts to kinetic energy as the ball fall down. $(\text { K.E. })_{\mathrm{i}}=\mathrm{mgh}$ $- When the ball finally hits the ground all the potential energy has convert to kinetic energy.$ $(\text { K.E. })_{\mathrm{f}}=\frac{\mathrm{mgh}}{4}$ Fraction of the energy loss $\Delta \mathrm{K}=\frac{(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}}{(\text { K.E. })_{\mathrm{i}}}$ $\Delta \mathrm{K} =\frac{\mathrm{mgh}-\frac{\mathrm{mgh}}{4}}{\mathrm{mgh}}$ $\Delta \mathrm{K} =\frac{3}{4}$
Kerala CEE 2021
Work, Energy and Power
149102
A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x \mathrm{~cm}$ into the sand, the average resistance offered by the sand to the body is
B Work energy theorem when body reaches sand floor Potential energy $=$ kinetic energy $\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^{2}$ $\mathrm{v}=\sqrt{2 \mathrm{gh}}$ When body penetrates to the sand it's KE will be dissipated by the resultant resistance force $\frac{1}{2} \mathrm{Mv}^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\frac{1}{2} \mathrm{M}(\sqrt{2 \mathrm{gh}})^{2}=(\mathrm{F}-\mathrm{Mg}) \mathrm{x}$ $\mathrm{Mgh}+\mathrm{Mgx}=\mathrm{Fx}$ $\mathrm{F}=\operatorname{Mg}\left(\frac{\mathrm{x}+\mathrm{h}}{\mathrm{x}}\right)$
