149019
Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
D Potential energy is conservative. Frictional force is non-conservative force as work done against frictional force can not be stored as potential energy. Hence, potential energy cannot be associated with frictional forces. Note- As we know, conservative force are those force which depend upon the initial and final position while non-conservative force depend upon the path.
AIIMS-2005
Work, Energy and Power
149020
Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s=0$ to $20 \mathrm{~m}$ will be
1 $250 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $150 \mathrm{~J}$
4 $10 \mathrm{~J}$
Explanation:
A Given, We know that, Loss of kinetic energy $=$ Work done by the frictional force Loss of kinetic energy $=$ Area under the $\mathrm{F}$ v/s S curve $(\text { K.E. })_{\text {loss }}=\left[\frac{(5+15)}{2} \times 5\right]+[15 \times 5]+\left[\frac{(15+10)}{2} \times 10\right]$ $(\text { K.E. })_{\text {loss }}=\left[\frac{20}{2} \times 5\right]+[15 \times 5]+\left[\frac{25}{2} \times 10\right]$ $\text { (K.E. })_{\text {loss }}=(10 \times 5)+75+(25 \times 5)$ $\text { (K.E. })_{\text {loss }}=(50+75+125) \mathrm{J}$ $\text { (K.E. })_{\text {loss }}=250 \mathrm{~J}$
AIIMS-2015
Work, Energy and Power
149021
The potential energy of a certain particle is given by $U=\frac{1}{2}\left(x^{2}-z^{2}\right)$. The force on it is:
1 $-x \hat{i}+z \hat{k}$
2 $x \hat{i}+z \hat{k}$
3 $\frac{1}{2}(x \hat{i}+z \hat{k})$
4 $\frac{1}{2}(x \hat{i}-z \hat{k})$
Explanation:
A Given, potential energy of a certain particle $(\mathrm{U})=\frac{1}{2}\left(\mathrm{x}^{2}-\mathrm{z}^{2}\right)$ Now, $\quad F_{x}=-\frac{d U}{d x}$ $F_{x}=-\frac{d}{d x}\left(\frac{x^{2}-z^{2}}{2}\right)$ $F_{x}=-\frac{1}{2}[2 x-0]$ $\begin{aligned} \mathrm{F}_{\mathrm{x}} =-\frac{2 \mathrm{x}}{2}=-\mathrm{x} \\ \text { And, } \quad \mathrm{F}_{\mathrm{z}} =-\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{\mathrm{x}^{2}-\mathrm{z}^{2}}{2}\right) \\ \mathrm{F}_{\mathrm{z}} =-\frac{1}{2}[0-2 \mathrm{z}] \\ \mathrm{F}_{\mathrm{z}} =\frac{2 \mathrm{z}}{2}=\mathrm{z} \\ \text { Therefore, } \overrightarrow{\mathrm{F}}=\mathrm{F}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{z}} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{F}} =-\mathrm{x} \hat{\mathrm{i}}+\mathrm{zk}\end{aligned}$
AIIMS-2011
Work, Energy and Power
149022
A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4 \mathrm{~m}$ to have it bounce back to the same height (ignore air resistance)?
1 $6.55 \mathrm{~m} / \mathrm{s}$
2 $12.0 \mathrm{~m} / \mathrm{s}$
3 $8.6 \mathrm{~m} / \mathrm{s}$
4 $4.55 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, height (h) $=12.4 \mathrm{~m}$ A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. From the equation of motion- $v^{2}=u^{2}+2 g h$ $v^{2}=u^{2}+(2 \times 10 \times 12.4)$ $v^{2}=u^{2}+(2 \times 124)$ $v^{2}=u^{2}+248$ Kinetic energy of the ball (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ A ball loss $15.0 \%$ of its kinetic energy $\text { K.E. }{ }^{\prime}=\left(\frac{100-15}{100}\right) \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\text { K.E. }{ }^{\prime}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ Let $\mathrm{v}_{1}$ be the upward velocity just after the collision with the ground. $\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\mathrm{v}_{1}^{2}=\frac{85}{100}\left(\mathrm{u}^{2}+248\right)$ Now, taking upward motion - $0= \frac{85}{100}\left(u^{2}+248\right)-2 \times 10 \times 12.4 \quad\left(v=0, u=v_{1}\right)$ $\frac{85}{100}\left(u^{2}+248\right)=248$ $u^{2}+248=\frac{248 \times 100}{85}$ $u^{2}= 291.76-248$ $u =\sqrt{43.76}=6.61 \simeq 6.55 \mathrm{~m} / \mathrm{s}$
AIIMS-2010
Work, Energy and Power
149024
A bullet of mass $10 \mathrm{~g}$ leaves a rifle at an initial velocity of $1000 \mathrm{~m} / \mathrm{sec}$ and strikes the earth at the same level with a velocity of $500 \mathrm{~m} / \mathrm{sec}$. The work in Joule to overcoming the resistance of air will be:
1 $500 \mathrm{~J}$
2 $5000 \mathrm{~J}$
3 $3750 \mathrm{~J}$
4 $475 \mathrm{~J}$
Explanation:
C Given, mass of bullet $(\mathrm{m})=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}=10^{-2} \mathrm{~kg}$, initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)=1000 \mathrm{~m} / \mathrm{sec}$, final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)=500 \mathrm{~m} / \mathrm{sec}$ Work done in overcoming the air resistance $=\Delta \mathrm{KE}$ Work done $(\mathrm{W})=(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{i}}^{2}-\mathrm{v}_{\mathrm{f}}^{2}\right)$ $\mathrm{W}=\frac{1}{2} \times 10^{-2}\left[(1000)^{2}-(500)^{2}\right]$ $\mathrm{W}=\frac{1}{2} \times \frac{1}{100}[1000000-250000]$ $\mathrm{W}=\frac{1}{200} \times 750000$ $\mathrm{~W}=\frac{7500}{2}$ $\mathrm{~W}=3750 \mathrm{~J}$
149019
Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
D Potential energy is conservative. Frictional force is non-conservative force as work done against frictional force can not be stored as potential energy. Hence, potential energy cannot be associated with frictional forces. Note- As we know, conservative force are those force which depend upon the initial and final position while non-conservative force depend upon the path.
AIIMS-2005
Work, Energy and Power
149020
Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s=0$ to $20 \mathrm{~m}$ will be
1 $250 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $150 \mathrm{~J}$
4 $10 \mathrm{~J}$
Explanation:
A Given, We know that, Loss of kinetic energy $=$ Work done by the frictional force Loss of kinetic energy $=$ Area under the $\mathrm{F}$ v/s S curve $(\text { K.E. })_{\text {loss }}=\left[\frac{(5+15)}{2} \times 5\right]+[15 \times 5]+\left[\frac{(15+10)}{2} \times 10\right]$ $(\text { K.E. })_{\text {loss }}=\left[\frac{20}{2} \times 5\right]+[15 \times 5]+\left[\frac{25}{2} \times 10\right]$ $\text { (K.E. })_{\text {loss }}=(10 \times 5)+75+(25 \times 5)$ $\text { (K.E. })_{\text {loss }}=(50+75+125) \mathrm{J}$ $\text { (K.E. })_{\text {loss }}=250 \mathrm{~J}$
AIIMS-2015
Work, Energy and Power
149021
The potential energy of a certain particle is given by $U=\frac{1}{2}\left(x^{2}-z^{2}\right)$. The force on it is:
1 $-x \hat{i}+z \hat{k}$
2 $x \hat{i}+z \hat{k}$
3 $\frac{1}{2}(x \hat{i}+z \hat{k})$
4 $\frac{1}{2}(x \hat{i}-z \hat{k})$
Explanation:
A Given, potential energy of a certain particle $(\mathrm{U})=\frac{1}{2}\left(\mathrm{x}^{2}-\mathrm{z}^{2}\right)$ Now, $\quad F_{x}=-\frac{d U}{d x}$ $F_{x}=-\frac{d}{d x}\left(\frac{x^{2}-z^{2}}{2}\right)$ $F_{x}=-\frac{1}{2}[2 x-0]$ $\begin{aligned} \mathrm{F}_{\mathrm{x}} =-\frac{2 \mathrm{x}}{2}=-\mathrm{x} \\ \text { And, } \quad \mathrm{F}_{\mathrm{z}} =-\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{\mathrm{x}^{2}-\mathrm{z}^{2}}{2}\right) \\ \mathrm{F}_{\mathrm{z}} =-\frac{1}{2}[0-2 \mathrm{z}] \\ \mathrm{F}_{\mathrm{z}} =\frac{2 \mathrm{z}}{2}=\mathrm{z} \\ \text { Therefore, } \overrightarrow{\mathrm{F}}=\mathrm{F}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{z}} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{F}} =-\mathrm{x} \hat{\mathrm{i}}+\mathrm{zk}\end{aligned}$
AIIMS-2011
Work, Energy and Power
149022
A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4 \mathrm{~m}$ to have it bounce back to the same height (ignore air resistance)?
1 $6.55 \mathrm{~m} / \mathrm{s}$
2 $12.0 \mathrm{~m} / \mathrm{s}$
3 $8.6 \mathrm{~m} / \mathrm{s}$
4 $4.55 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, height (h) $=12.4 \mathrm{~m}$ A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. From the equation of motion- $v^{2}=u^{2}+2 g h$ $v^{2}=u^{2}+(2 \times 10 \times 12.4)$ $v^{2}=u^{2}+(2 \times 124)$ $v^{2}=u^{2}+248$ Kinetic energy of the ball (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ A ball loss $15.0 \%$ of its kinetic energy $\text { K.E. }{ }^{\prime}=\left(\frac{100-15}{100}\right) \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\text { K.E. }{ }^{\prime}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ Let $\mathrm{v}_{1}$ be the upward velocity just after the collision with the ground. $\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\mathrm{v}_{1}^{2}=\frac{85}{100}\left(\mathrm{u}^{2}+248\right)$ Now, taking upward motion - $0= \frac{85}{100}\left(u^{2}+248\right)-2 \times 10 \times 12.4 \quad\left(v=0, u=v_{1}\right)$ $\frac{85}{100}\left(u^{2}+248\right)=248$ $u^{2}+248=\frac{248 \times 100}{85}$ $u^{2}= 291.76-248$ $u =\sqrt{43.76}=6.61 \simeq 6.55 \mathrm{~m} / \mathrm{s}$
AIIMS-2010
Work, Energy and Power
149024
A bullet of mass $10 \mathrm{~g}$ leaves a rifle at an initial velocity of $1000 \mathrm{~m} / \mathrm{sec}$ and strikes the earth at the same level with a velocity of $500 \mathrm{~m} / \mathrm{sec}$. The work in Joule to overcoming the resistance of air will be:
1 $500 \mathrm{~J}$
2 $5000 \mathrm{~J}$
3 $3750 \mathrm{~J}$
4 $475 \mathrm{~J}$
Explanation:
C Given, mass of bullet $(\mathrm{m})=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}=10^{-2} \mathrm{~kg}$, initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)=1000 \mathrm{~m} / \mathrm{sec}$, final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)=500 \mathrm{~m} / \mathrm{sec}$ Work done in overcoming the air resistance $=\Delta \mathrm{KE}$ Work done $(\mathrm{W})=(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{i}}^{2}-\mathrm{v}_{\mathrm{f}}^{2}\right)$ $\mathrm{W}=\frac{1}{2} \times 10^{-2}\left[(1000)^{2}-(500)^{2}\right]$ $\mathrm{W}=\frac{1}{2} \times \frac{1}{100}[1000000-250000]$ $\mathrm{W}=\frac{1}{200} \times 750000$ $\mathrm{~W}=\frac{7500}{2}$ $\mathrm{~W}=3750 \mathrm{~J}$
149019
Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
D Potential energy is conservative. Frictional force is non-conservative force as work done against frictional force can not be stored as potential energy. Hence, potential energy cannot be associated with frictional forces. Note- As we know, conservative force are those force which depend upon the initial and final position while non-conservative force depend upon the path.
AIIMS-2005
Work, Energy and Power
149020
Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s=0$ to $20 \mathrm{~m}$ will be
1 $250 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $150 \mathrm{~J}$
4 $10 \mathrm{~J}$
Explanation:
A Given, We know that, Loss of kinetic energy $=$ Work done by the frictional force Loss of kinetic energy $=$ Area under the $\mathrm{F}$ v/s S curve $(\text { K.E. })_{\text {loss }}=\left[\frac{(5+15)}{2} \times 5\right]+[15 \times 5]+\left[\frac{(15+10)}{2} \times 10\right]$ $(\text { K.E. })_{\text {loss }}=\left[\frac{20}{2} \times 5\right]+[15 \times 5]+\left[\frac{25}{2} \times 10\right]$ $\text { (K.E. })_{\text {loss }}=(10 \times 5)+75+(25 \times 5)$ $\text { (K.E. })_{\text {loss }}=(50+75+125) \mathrm{J}$ $\text { (K.E. })_{\text {loss }}=250 \mathrm{~J}$
AIIMS-2015
Work, Energy and Power
149021
The potential energy of a certain particle is given by $U=\frac{1}{2}\left(x^{2}-z^{2}\right)$. The force on it is:
1 $-x \hat{i}+z \hat{k}$
2 $x \hat{i}+z \hat{k}$
3 $\frac{1}{2}(x \hat{i}+z \hat{k})$
4 $\frac{1}{2}(x \hat{i}-z \hat{k})$
Explanation:
A Given, potential energy of a certain particle $(\mathrm{U})=\frac{1}{2}\left(\mathrm{x}^{2}-\mathrm{z}^{2}\right)$ Now, $\quad F_{x}=-\frac{d U}{d x}$ $F_{x}=-\frac{d}{d x}\left(\frac{x^{2}-z^{2}}{2}\right)$ $F_{x}=-\frac{1}{2}[2 x-0]$ $\begin{aligned} \mathrm{F}_{\mathrm{x}} =-\frac{2 \mathrm{x}}{2}=-\mathrm{x} \\ \text { And, } \quad \mathrm{F}_{\mathrm{z}} =-\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{\mathrm{x}^{2}-\mathrm{z}^{2}}{2}\right) \\ \mathrm{F}_{\mathrm{z}} =-\frac{1}{2}[0-2 \mathrm{z}] \\ \mathrm{F}_{\mathrm{z}} =\frac{2 \mathrm{z}}{2}=\mathrm{z} \\ \text { Therefore, } \overrightarrow{\mathrm{F}}=\mathrm{F}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{z}} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{F}} =-\mathrm{x} \hat{\mathrm{i}}+\mathrm{zk}\end{aligned}$
AIIMS-2011
Work, Energy and Power
149022
A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4 \mathrm{~m}$ to have it bounce back to the same height (ignore air resistance)?
1 $6.55 \mathrm{~m} / \mathrm{s}$
2 $12.0 \mathrm{~m} / \mathrm{s}$
3 $8.6 \mathrm{~m} / \mathrm{s}$
4 $4.55 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, height (h) $=12.4 \mathrm{~m}$ A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. From the equation of motion- $v^{2}=u^{2}+2 g h$ $v^{2}=u^{2}+(2 \times 10 \times 12.4)$ $v^{2}=u^{2}+(2 \times 124)$ $v^{2}=u^{2}+248$ Kinetic energy of the ball (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ A ball loss $15.0 \%$ of its kinetic energy $\text { K.E. }{ }^{\prime}=\left(\frac{100-15}{100}\right) \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\text { K.E. }{ }^{\prime}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ Let $\mathrm{v}_{1}$ be the upward velocity just after the collision with the ground. $\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\mathrm{v}_{1}^{2}=\frac{85}{100}\left(\mathrm{u}^{2}+248\right)$ Now, taking upward motion - $0= \frac{85}{100}\left(u^{2}+248\right)-2 \times 10 \times 12.4 \quad\left(v=0, u=v_{1}\right)$ $\frac{85}{100}\left(u^{2}+248\right)=248$ $u^{2}+248=\frac{248 \times 100}{85}$ $u^{2}= 291.76-248$ $u =\sqrt{43.76}=6.61 \simeq 6.55 \mathrm{~m} / \mathrm{s}$
AIIMS-2010
Work, Energy and Power
149024
A bullet of mass $10 \mathrm{~g}$ leaves a rifle at an initial velocity of $1000 \mathrm{~m} / \mathrm{sec}$ and strikes the earth at the same level with a velocity of $500 \mathrm{~m} / \mathrm{sec}$. The work in Joule to overcoming the resistance of air will be:
1 $500 \mathrm{~J}$
2 $5000 \mathrm{~J}$
3 $3750 \mathrm{~J}$
4 $475 \mathrm{~J}$
Explanation:
C Given, mass of bullet $(\mathrm{m})=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}=10^{-2} \mathrm{~kg}$, initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)=1000 \mathrm{~m} / \mathrm{sec}$, final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)=500 \mathrm{~m} / \mathrm{sec}$ Work done in overcoming the air resistance $=\Delta \mathrm{KE}$ Work done $(\mathrm{W})=(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{i}}^{2}-\mathrm{v}_{\mathrm{f}}^{2}\right)$ $\mathrm{W}=\frac{1}{2} \times 10^{-2}\left[(1000)^{2}-(500)^{2}\right]$ $\mathrm{W}=\frac{1}{2} \times \frac{1}{100}[1000000-250000]$ $\mathrm{W}=\frac{1}{200} \times 750000$ $\mathrm{~W}=\frac{7500}{2}$ $\mathrm{~W}=3750 \mathrm{~J}$
149019
Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
D Potential energy is conservative. Frictional force is non-conservative force as work done against frictional force can not be stored as potential energy. Hence, potential energy cannot be associated with frictional forces. Note- As we know, conservative force are those force which depend upon the initial and final position while non-conservative force depend upon the path.
AIIMS-2005
Work, Energy and Power
149020
Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s=0$ to $20 \mathrm{~m}$ will be
1 $250 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $150 \mathrm{~J}$
4 $10 \mathrm{~J}$
Explanation:
A Given, We know that, Loss of kinetic energy $=$ Work done by the frictional force Loss of kinetic energy $=$ Area under the $\mathrm{F}$ v/s S curve $(\text { K.E. })_{\text {loss }}=\left[\frac{(5+15)}{2} \times 5\right]+[15 \times 5]+\left[\frac{(15+10)}{2} \times 10\right]$ $(\text { K.E. })_{\text {loss }}=\left[\frac{20}{2} \times 5\right]+[15 \times 5]+\left[\frac{25}{2} \times 10\right]$ $\text { (K.E. })_{\text {loss }}=(10 \times 5)+75+(25 \times 5)$ $\text { (K.E. })_{\text {loss }}=(50+75+125) \mathrm{J}$ $\text { (K.E. })_{\text {loss }}=250 \mathrm{~J}$
AIIMS-2015
Work, Energy and Power
149021
The potential energy of a certain particle is given by $U=\frac{1}{2}\left(x^{2}-z^{2}\right)$. The force on it is:
1 $-x \hat{i}+z \hat{k}$
2 $x \hat{i}+z \hat{k}$
3 $\frac{1}{2}(x \hat{i}+z \hat{k})$
4 $\frac{1}{2}(x \hat{i}-z \hat{k})$
Explanation:
A Given, potential energy of a certain particle $(\mathrm{U})=\frac{1}{2}\left(\mathrm{x}^{2}-\mathrm{z}^{2}\right)$ Now, $\quad F_{x}=-\frac{d U}{d x}$ $F_{x}=-\frac{d}{d x}\left(\frac{x^{2}-z^{2}}{2}\right)$ $F_{x}=-\frac{1}{2}[2 x-0]$ $\begin{aligned} \mathrm{F}_{\mathrm{x}} =-\frac{2 \mathrm{x}}{2}=-\mathrm{x} \\ \text { And, } \quad \mathrm{F}_{\mathrm{z}} =-\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{\mathrm{x}^{2}-\mathrm{z}^{2}}{2}\right) \\ \mathrm{F}_{\mathrm{z}} =-\frac{1}{2}[0-2 \mathrm{z}] \\ \mathrm{F}_{\mathrm{z}} =\frac{2 \mathrm{z}}{2}=\mathrm{z} \\ \text { Therefore, } \overrightarrow{\mathrm{F}}=\mathrm{F}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{z}} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{F}} =-\mathrm{x} \hat{\mathrm{i}}+\mathrm{zk}\end{aligned}$
AIIMS-2011
Work, Energy and Power
149022
A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4 \mathrm{~m}$ to have it bounce back to the same height (ignore air resistance)?
1 $6.55 \mathrm{~m} / \mathrm{s}$
2 $12.0 \mathrm{~m} / \mathrm{s}$
3 $8.6 \mathrm{~m} / \mathrm{s}$
4 $4.55 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, height (h) $=12.4 \mathrm{~m}$ A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. From the equation of motion- $v^{2}=u^{2}+2 g h$ $v^{2}=u^{2}+(2 \times 10 \times 12.4)$ $v^{2}=u^{2}+(2 \times 124)$ $v^{2}=u^{2}+248$ Kinetic energy of the ball (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ A ball loss $15.0 \%$ of its kinetic energy $\text { K.E. }{ }^{\prime}=\left(\frac{100-15}{100}\right) \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\text { K.E. }{ }^{\prime}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ Let $\mathrm{v}_{1}$ be the upward velocity just after the collision with the ground. $\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\mathrm{v}_{1}^{2}=\frac{85}{100}\left(\mathrm{u}^{2}+248\right)$ Now, taking upward motion - $0= \frac{85}{100}\left(u^{2}+248\right)-2 \times 10 \times 12.4 \quad\left(v=0, u=v_{1}\right)$ $\frac{85}{100}\left(u^{2}+248\right)=248$ $u^{2}+248=\frac{248 \times 100}{85}$ $u^{2}= 291.76-248$ $u =\sqrt{43.76}=6.61 \simeq 6.55 \mathrm{~m} / \mathrm{s}$
AIIMS-2010
Work, Energy and Power
149024
A bullet of mass $10 \mathrm{~g}$ leaves a rifle at an initial velocity of $1000 \mathrm{~m} / \mathrm{sec}$ and strikes the earth at the same level with a velocity of $500 \mathrm{~m} / \mathrm{sec}$. The work in Joule to overcoming the resistance of air will be:
1 $500 \mathrm{~J}$
2 $5000 \mathrm{~J}$
3 $3750 \mathrm{~J}$
4 $475 \mathrm{~J}$
Explanation:
C Given, mass of bullet $(\mathrm{m})=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}=10^{-2} \mathrm{~kg}$, initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)=1000 \mathrm{~m} / \mathrm{sec}$, final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)=500 \mathrm{~m} / \mathrm{sec}$ Work done in overcoming the air resistance $=\Delta \mathrm{KE}$ Work done $(\mathrm{W})=(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{i}}^{2}-\mathrm{v}_{\mathrm{f}}^{2}\right)$ $\mathrm{W}=\frac{1}{2} \times 10^{-2}\left[(1000)^{2}-(500)^{2}\right]$ $\mathrm{W}=\frac{1}{2} \times \frac{1}{100}[1000000-250000]$ $\mathrm{W}=\frac{1}{200} \times 750000$ $\mathrm{~W}=\frac{7500}{2}$ $\mathrm{~W}=3750 \mathrm{~J}$
149019
Assertion: Frictional forces are conservative forces. Reason: Potential energy can be associated with frictional forces.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
D Potential energy is conservative. Frictional force is non-conservative force as work done against frictional force can not be stored as potential energy. Hence, potential energy cannot be associated with frictional forces. Note- As we know, conservative force are those force which depend upon the initial and final position while non-conservative force depend upon the path.
AIIMS-2005
Work, Energy and Power
149020
Figure here shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s=0$ to $20 \mathrm{~m}$ will be
1 $250 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $150 \mathrm{~J}$
4 $10 \mathrm{~J}$
Explanation:
A Given, We know that, Loss of kinetic energy $=$ Work done by the frictional force Loss of kinetic energy $=$ Area under the $\mathrm{F}$ v/s S curve $(\text { K.E. })_{\text {loss }}=\left[\frac{(5+15)}{2} \times 5\right]+[15 \times 5]+\left[\frac{(15+10)}{2} \times 10\right]$ $(\text { K.E. })_{\text {loss }}=\left[\frac{20}{2} \times 5\right]+[15 \times 5]+\left[\frac{25}{2} \times 10\right]$ $\text { (K.E. })_{\text {loss }}=(10 \times 5)+75+(25 \times 5)$ $\text { (K.E. })_{\text {loss }}=(50+75+125) \mathrm{J}$ $\text { (K.E. })_{\text {loss }}=250 \mathrm{~J}$
AIIMS-2015
Work, Energy and Power
149021
The potential energy of a certain particle is given by $U=\frac{1}{2}\left(x^{2}-z^{2}\right)$. The force on it is:
1 $-x \hat{i}+z \hat{k}$
2 $x \hat{i}+z \hat{k}$
3 $\frac{1}{2}(x \hat{i}+z \hat{k})$
4 $\frac{1}{2}(x \hat{i}-z \hat{k})$
Explanation:
A Given, potential energy of a certain particle $(\mathrm{U})=\frac{1}{2}\left(\mathrm{x}^{2}-\mathrm{z}^{2}\right)$ Now, $\quad F_{x}=-\frac{d U}{d x}$ $F_{x}=-\frac{d}{d x}\left(\frac{x^{2}-z^{2}}{2}\right)$ $F_{x}=-\frac{1}{2}[2 x-0]$ $\begin{aligned} \mathrm{F}_{\mathrm{x}} =-\frac{2 \mathrm{x}}{2}=-\mathrm{x} \\ \text { And, } \quad \mathrm{F}_{\mathrm{z}} =-\frac{\mathrm{d}}{\mathrm{dz}}\left(\frac{\mathrm{x}^{2}-\mathrm{z}^{2}}{2}\right) \\ \mathrm{F}_{\mathrm{z}} =-\frac{1}{2}[0-2 \mathrm{z}] \\ \mathrm{F}_{\mathrm{z}} =\frac{2 \mathrm{z}}{2}=\mathrm{z} \\ \text { Therefore, } \overrightarrow{\mathrm{F}}=\mathrm{F}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{F}_{\mathrm{z}} \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{F}} =-\mathrm{x} \hat{\mathrm{i}}+\mathrm{zk}\end{aligned}$
AIIMS-2011
Work, Energy and Power
149022
A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4 \mathrm{~m}$ to have it bounce back to the same height (ignore air resistance)?
1 $6.55 \mathrm{~m} / \mathrm{s}$
2 $12.0 \mathrm{~m} / \mathrm{s}$
3 $8.6 \mathrm{~m} / \mathrm{s}$
4 $4.55 \mathrm{~m} / \mathrm{s}$
Explanation:
A Given, height (h) $=12.4 \mathrm{~m}$ A ball loses $15.0 \%$ of its kinetic energy when it bounces back from a concrete wall. From the equation of motion- $v^{2}=u^{2}+2 g h$ $v^{2}=u^{2}+(2 \times 10 \times 12.4)$ $v^{2}=u^{2}+(2 \times 124)$ $v^{2}=u^{2}+248$ Kinetic energy of the ball (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ $=\frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ A ball loss $15.0 \%$ of its kinetic energy $\text { K.E. }{ }^{\prime}=\left(\frac{100-15}{100}\right) \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\text { K.E. }{ }^{\prime}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ Let $\mathrm{v}_{1}$ be the upward velocity just after the collision with the ground. $\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{85}{100} \times \frac{1}{2} \mathrm{~m}\left(\mathrm{u}^{2}+248\right)$ $\mathrm{v}_{1}^{2}=\frac{85}{100}\left(\mathrm{u}^{2}+248\right)$ Now, taking upward motion - $0= \frac{85}{100}\left(u^{2}+248\right)-2 \times 10 \times 12.4 \quad\left(v=0, u=v_{1}\right)$ $\frac{85}{100}\left(u^{2}+248\right)=248$ $u^{2}+248=\frac{248 \times 100}{85}$ $u^{2}= 291.76-248$ $u =\sqrt{43.76}=6.61 \simeq 6.55 \mathrm{~m} / \mathrm{s}$
AIIMS-2010
Work, Energy and Power
149024
A bullet of mass $10 \mathrm{~g}$ leaves a rifle at an initial velocity of $1000 \mathrm{~m} / \mathrm{sec}$ and strikes the earth at the same level with a velocity of $500 \mathrm{~m} / \mathrm{sec}$. The work in Joule to overcoming the resistance of air will be:
1 $500 \mathrm{~J}$
2 $5000 \mathrm{~J}$
3 $3750 \mathrm{~J}$
4 $475 \mathrm{~J}$
Explanation:
C Given, mass of bullet $(\mathrm{m})=10 \mathrm{~g}=10 \times 10^{-3}$ $\mathrm{kg}=10^{-2} \mathrm{~kg}$, initial velocity $\left(\mathrm{v}_{\mathrm{i}}\right)=1000 \mathrm{~m} / \mathrm{sec}$, final velocity $\left(\mathrm{v}_{\mathrm{f}}\right)=500 \mathrm{~m} / \mathrm{sec}$ Work done in overcoming the air resistance $=\Delta \mathrm{KE}$ Work done $(\mathrm{W})=(\text { K.E. })_{\mathrm{i}}-(\text { K.E. })_{\mathrm{f}}$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{i}}^{2}-\mathrm{v}_{\mathrm{f}}^{2}\right)$ $\mathrm{W}=\frac{1}{2} \times 10^{-2}\left[(1000)^{2}-(500)^{2}\right]$ $\mathrm{W}=\frac{1}{2} \times \frac{1}{100}[1000000-250000]$ $\mathrm{W}=\frac{1}{200} \times 750000$ $\mathrm{~W}=\frac{7500}{2}$ $\mathrm{~W}=3750 \mathrm{~J}$
