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PHXI06:WORK ENERGY AND POWER

355589 The potential energy of a particle of mass 5 $k g$ moving in the $x y -$ plane is given by $U = (- 7 x + 24 y) J, x$ and $y$ being in metre. If the particle starts from rest and from origin, then speed of particle at $t = 2 s$ is

1

5 m s^{- 1}

2

0.1 m s^{- 1}

3

17.5 m s^{- 1}

4

10 m s^{- 1}

Explanation:

$F = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j} = 7 \hat{i} - 24 \hat{j}$
$∴ a_{x} = \frac{F_{x}}{m} = \frac{7}{5} = 1.4 m s^{- 2}$ along positive $x$ -axis
$a_{y} = \frac{F_{Y}}{m} = - \frac{24}{5}$
$= - 4.8 m s^{- 2}$ along negative $y$ -axis
$∴ v_{X} = a_{x} t = 1.4 \times 2 = 2.8 m s^{- 1}$
and $v_{Y} = 4.8 \times 2 = 9.6 m s^{- 1}$
$∴ v = \sqrt{v_{X}^{2} + v_{Y}^{2}} = 10 m s^{- 1}$

PHXI06:WORK ENERGY AND POWER

355590 A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r^{2}}$ . Its total energy is :

1

\frac{k}{2 a^{2}}

2 Zero

3

- \frac{3}{2} \frac{k}{a^{2}}

4

- \frac{k}{4 a^{2}}

Explanation:

$F = - \frac{\partial v}{\partial r} = \frac{K}{r^{3}}$
Since it is performing circular motion
$\begin{aligned} F = \frac{m v^{2}}{r} = \frac{K}{r^{3}} \\ m v^{2} = \frac{K}{r^{2}} \\ \Rightarrow K \cdot E = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} \end{aligned}$
Total energy $= P . E . + K \cdot E$
$= - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = 0$

PHXI06:WORK ENERGY AND POWER

355591 The potential energy for a force field $\vec{F}$ is given by $U (x, y) = \cos (x + y)$ . The force acting on a particle at position given by coordinates $(0, π / 4)$ is

1

- \frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

2

\frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

3

(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})

4

(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})

Explanation:

$F_{x} = - \frac{\partial U}{\partial x} = \sin (x + y),$
$F_{y} = - \frac{\partial U}{\partial x} = \sin (x + y)$
${F_{x} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}},$
${F_{y} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}}$
$F = \frac{1}{\sqrt{2}} [\hat{i} + \hat{j}]$

PHXI06:WORK ENERGY AND POWER

355592 The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particle $A$ is fixed at origin of a fixed coordinate system. A particle $B$ which is free to move experiences a force $\vec{F} = (- \frac{2 α}{r^{3}} + \frac{β}{r^{2}}) \hat{r}$ due to particle $A$ where $\hat{r}$ is the position vector of particle $B$ relative to $A$ . It is given that the force is conservative in nature and potential energy at infinity is zero. If $B$ has to be removed from the influence of $A$ , energy has to be supplied for such a process. The ionisation energy $E_{o}$ is at work that has to be done by an external agent to move the particle from a distance $r_{o}$ to infinity slowly. Here $r_{o}$ is the equilibrium position of the particle.
What is potential energy function of particle as function of $r$ ?

1

\frac{α}{r^{2}} - \frac{β}{r}

2

- \frac{α}{r^{2}} + \frac{β}{r}

3

- \frac{α}{r^{2}} - \frac{β}{r}

4

\frac{α}{r^{2}} + \frac{β}{r}

Explanation:

$F = \frac{- 2 α}{r^{3}} + \frac{β}{r^{2}} = \frac{- d U}{d r}$
$d U = \frac{2 α}{r^{3}} d r - \frac{β}{r^{2}} d r$
$\int_{0}^{U} d U = 2 α \int_{\infty}^{r} \frac{1}{r^{3}} d r - β \int_{\infty}^{r} \frac{1}{r^{2}} d r$
$U = \frac{- α}{r^{2}} + \frac{β}{r}$

PHXI06:WORK ENERGY AND POWER

355589 The potential energy of a particle of mass 5 $k g$ moving in the $x y -$ plane is given by $U = (- 7 x + 24 y) J, x$ and $y$ being in metre. If the particle starts from rest and from origin, then speed of particle at $t = 2 s$ is

1

5 m s^{- 1}

2

0.1 m s^{- 1}

3

17.5 m s^{- 1}

4

10 m s^{- 1}

Explanation:

$F = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j} = 7 \hat{i} - 24 \hat{j}$
$∴ a_{x} = \frac{F_{x}}{m} = \frac{7}{5} = 1.4 m s^{- 2}$ along positive $x$ -axis
$a_{y} = \frac{F_{Y}}{m} = - \frac{24}{5}$
$= - 4.8 m s^{- 2}$ along negative $y$ -axis
$∴ v_{X} = a_{x} t = 1.4 \times 2 = 2.8 m s^{- 1}$
and $v_{Y} = 4.8 \times 2 = 9.6 m s^{- 1}$
$∴ v = \sqrt{v_{X}^{2} + v_{Y}^{2}} = 10 m s^{- 1}$

PHXI06:WORK ENERGY AND POWER

355590 A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r^{2}}$ . Its total energy is :

1

\frac{k}{2 a^{2}}

2 Zero

3

- \frac{3}{2} \frac{k}{a^{2}}

4

- \frac{k}{4 a^{2}}

Explanation:

$F = - \frac{\partial v}{\partial r} = \frac{K}{r^{3}}$
Since it is performing circular motion
$\begin{aligned} F = \frac{m v^{2}}{r} = \frac{K}{r^{3}} \\ m v^{2} = \frac{K}{r^{2}} \\ \Rightarrow K \cdot E = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} \end{aligned}$
Total energy $= P . E . + K \cdot E$
$= - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = 0$

PHXI06:WORK ENERGY AND POWER

355591 The potential energy for a force field $\vec{F}$ is given by $U (x, y) = \cos (x + y)$ . The force acting on a particle at position given by coordinates $(0, π / 4)$ is

1

- \frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

2

\frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

3

(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})

4

(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})

Explanation:

$F_{x} = - \frac{\partial U}{\partial x} = \sin (x + y),$
$F_{y} = - \frac{\partial U}{\partial x} = \sin (x + y)$
${F_{x} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}},$
${F_{y} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}}$
$F = \frac{1}{\sqrt{2}} [\hat{i} + \hat{j}]$

PHXI06:WORK ENERGY AND POWER

355592 The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particle $A$ is fixed at origin of a fixed coordinate system. A particle $B$ which is free to move experiences a force $\vec{F} = (- \frac{2 α}{r^{3}} + \frac{β}{r^{2}}) \hat{r}$ due to particle $A$ where $\hat{r}$ is the position vector of particle $B$ relative to $A$ . It is given that the force is conservative in nature and potential energy at infinity is zero. If $B$ has to be removed from the influence of $A$ , energy has to be supplied for such a process. The ionisation energy $E_{o}$ is at work that has to be done by an external agent to move the particle from a distance $r_{o}$ to infinity slowly. Here $r_{o}$ is the equilibrium position of the particle.
What is potential energy function of particle as function of $r$ ?

1

\frac{α}{r^{2}} - \frac{β}{r}

2

- \frac{α}{r^{2}} + \frac{β}{r}

3

- \frac{α}{r^{2}} - \frac{β}{r}

4

\frac{α}{r^{2}} + \frac{β}{r}

Explanation:

$F = \frac{- 2 α}{r^{3}} + \frac{β}{r^{2}} = \frac{- d U}{d r}$
$d U = \frac{2 α}{r^{3}} d r - \frac{β}{r^{2}} d r$
$\int_{0}^{U} d U = 2 α \int_{\infty}^{r} \frac{1}{r^{3}} d r - β \int_{\infty}^{r} \frac{1}{r^{2}} d r$
$U = \frac{- α}{r^{2}} + \frac{β}{r}$

PHXI06:WORK ENERGY AND POWER

355589 The potential energy of a particle of mass 5 $k g$ moving in the $x y -$ plane is given by $U = (- 7 x + 24 y) J, x$ and $y$ being in metre. If the particle starts from rest and from origin, then speed of particle at $t = 2 s$ is

1

5 m s^{- 1}

2

0.1 m s^{- 1}

3

17.5 m s^{- 1}

4

10 m s^{- 1}

Explanation:

$F = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j} = 7 \hat{i} - 24 \hat{j}$
$∴ a_{x} = \frac{F_{x}}{m} = \frac{7}{5} = 1.4 m s^{- 2}$ along positive $x$ -axis
$a_{y} = \frac{F_{Y}}{m} = - \frac{24}{5}$
$= - 4.8 m s^{- 2}$ along negative $y$ -axis
$∴ v_{X} = a_{x} t = 1.4 \times 2 = 2.8 m s^{- 1}$
and $v_{Y} = 4.8 \times 2 = 9.6 m s^{- 1}$
$∴ v = \sqrt{v_{X}^{2} + v_{Y}^{2}} = 10 m s^{- 1}$

PHXI06:WORK ENERGY AND POWER

355590 A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r^{2}}$ . Its total energy is :

1

\frac{k}{2 a^{2}}

2 Zero

3

- \frac{3}{2} \frac{k}{a^{2}}

4

- \frac{k}{4 a^{2}}

Explanation:

$F = - \frac{\partial v}{\partial r} = \frac{K}{r^{3}}$
Since it is performing circular motion
$\begin{aligned} F = \frac{m v^{2}}{r} = \frac{K}{r^{3}} \\ m v^{2} = \frac{K}{r^{2}} \\ \Rightarrow K \cdot E = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} \end{aligned}$
Total energy $= P . E . + K \cdot E$
$= - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = 0$

PHXI06:WORK ENERGY AND POWER

355591 The potential energy for a force field $\vec{F}$ is given by $U (x, y) = \cos (x + y)$ . The force acting on a particle at position given by coordinates $(0, π / 4)$ is

1

- \frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

2

\frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

3

(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})

4

(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})

Explanation:

$F_{x} = - \frac{\partial U}{\partial x} = \sin (x + y),$
$F_{y} = - \frac{\partial U}{\partial x} = \sin (x + y)$
${F_{x} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}},$
${F_{y} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}}$
$F = \frac{1}{\sqrt{2}} [\hat{i} + \hat{j}]$

PHXI06:WORK ENERGY AND POWER

355592 The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particle $A$ is fixed at origin of a fixed coordinate system. A particle $B$ which is free to move experiences a force $\vec{F} = (- \frac{2 α}{r^{3}} + \frac{β}{r^{2}}) \hat{r}$ due to particle $A$ where $\hat{r}$ is the position vector of particle $B$ relative to $A$ . It is given that the force is conservative in nature and potential energy at infinity is zero. If $B$ has to be removed from the influence of $A$ , energy has to be supplied for such a process. The ionisation energy $E_{o}$ is at work that has to be done by an external agent to move the particle from a distance $r_{o}$ to infinity slowly. Here $r_{o}$ is the equilibrium position of the particle.
What is potential energy function of particle as function of $r$ ?

1

\frac{α}{r^{2}} - \frac{β}{r}

2

- \frac{α}{r^{2}} + \frac{β}{r}

3

- \frac{α}{r^{2}} - \frac{β}{r}

4

\frac{α}{r^{2}} + \frac{β}{r}

Explanation:

$F = \frac{- 2 α}{r^{3}} + \frac{β}{r^{2}} = \frac{- d U}{d r}$
$d U = \frac{2 α}{r^{3}} d r - \frac{β}{r^{2}} d r$
$\int_{0}^{U} d U = 2 α \int_{\infty}^{r} \frac{1}{r^{3}} d r - β \int_{\infty}^{r} \frac{1}{r^{2}} d r$
$U = \frac{- α}{r^{2}} + \frac{β}{r}$

PHXI06:WORK ENERGY AND POWER

355589 The potential energy of a particle of mass 5 $k g$ moving in the $x y -$ plane is given by $U = (- 7 x + 24 y) J, x$ and $y$ being in metre. If the particle starts from rest and from origin, then speed of particle at $t = 2 s$ is

1

5 m s^{- 1}

2

0.1 m s^{- 1}

3

17.5 m s^{- 1}

4

10 m s^{- 1}

Explanation:

$F = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j} = 7 \hat{i} - 24 \hat{j}$
$∴ a_{x} = \frac{F_{x}}{m} = \frac{7}{5} = 1.4 m s^{- 2}$ along positive $x$ -axis
$a_{y} = \frac{F_{Y}}{m} = - \frac{24}{5}$
$= - 4.8 m s^{- 2}$ along negative $y$ -axis
$∴ v_{X} = a_{x} t = 1.4 \times 2 = 2.8 m s^{- 1}$
and $v_{Y} = 4.8 \times 2 = 9.6 m s^{- 1}$
$∴ v = \sqrt{v_{X}^{2} + v_{Y}^{2}} = 10 m s^{- 1}$

PHXI06:WORK ENERGY AND POWER

355590 A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U = - \frac{k}{2 r^{2}}$ . Its total energy is :

1

\frac{k}{2 a^{2}}

2 Zero

3

- \frac{3}{2} \frac{k}{a^{2}}

4

- \frac{k}{4 a^{2}}

Explanation:

$F = - \frac{\partial v}{\partial r} = \frac{K}{r^{3}}$
Since it is performing circular motion
$\begin{aligned} F = \frac{m v^{2}}{r} = \frac{K}{r^{3}} \\ m v^{2} = \frac{K}{r^{2}} \\ \Rightarrow K \cdot E = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} \end{aligned}$
Total energy $= P . E . + K \cdot E$
$= - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = 0$

PHXI06:WORK ENERGY AND POWER

355591 The potential energy for a force field $\vec{F}$ is given by $U (x, y) = \cos (x + y)$ . The force acting on a particle at position given by coordinates $(0, π / 4)$ is

1

- \frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

2

\frac{1}{\sqrt{2}} (\hat{i} + \hat{j})

3

(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j})

4

(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j})

Explanation:

$F_{x} = - \frac{\partial U}{\partial x} = \sin (x + y),$
$F_{y} = - \frac{\partial U}{\partial x} = \sin (x + y)$
${F_{x} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}},$
${F_{y} = \sin (x + y)]}_{(0, π / 4)} = \frac{1}{\sqrt{2}}$
$F = \frac{1}{\sqrt{2}} [\hat{i} + \hat{j}]$

PHXI06:WORK ENERGY AND POWER

355592 The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particle $A$ is fixed at origin of a fixed coordinate system. A particle $B$ which is free to move experiences a force $\vec{F} = (- \frac{2 α}{r^{3}} + \frac{β}{r^{2}}) \hat{r}$ due to particle $A$ where $\hat{r}$ is the position vector of particle $B$ relative to $A$ . It is given that the force is conservative in nature and potential energy at infinity is zero. If $B$ has to be removed from the influence of $A$ , energy has to be supplied for such a process. The ionisation energy $E_{o}$ is at work that has to be done by an external agent to move the particle from a distance $r_{o}$ to infinity slowly. Here $r_{o}$ is the equilibrium position of the particle.
What is potential energy function of particle as function of $r$ ?

1

\frac{α}{r^{2}} - \frac{β}{r}

2

- \frac{α}{r^{2}} + \frac{β}{r}

3

- \frac{α}{r^{2}} - \frac{β}{r}

4

\frac{α}{r^{2}} + \frac{β}{r}

Explanation:

$F = \frac{- 2 α}{r^{3}} + \frac{β}{r^{2}} = \frac{- d U}{d r}$
$d U = \frac{2 α}{r^{3}} d r - \frac{β}{r^{2}} d r$
$\int_{0}^{U} d U = 2 α \int_{\infty}^{r} \frac{1}{r^{3}} d r - β \int_{\infty}^{r} \frac{1}{r^{2}} d r$
$U = \frac{- α}{r^{2}} + \frac{β}{r}$

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