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PHXI06:WORK ENERGY AND POWER

355843 A body of mass $m$ starts moving from rest along $x$ -axis so that its velocity varies as $v = a \sqrt{s}$ where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

1

\frac{1}{4} m a t^{2}

2

4 m a^{4} t^{2}

3

\frac{1}{8} m a^{4} t^{2}

4

8 m a^{4} t^{2}

Explanation:

$v = a \sqrt{s} = \frac{d s}{d t}$
$\int_{0}^{s} \frac{d s}{\sqrt{s}} = a \int_{0}^{t} d t$
$2 \sqrt{s} = a t$
$v = \frac{a^{2} t}{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{8} m a^{4} t^{2}$

JEE - 2018

PHXI06:WORK ENERGY AND POWER

355844 A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$ , the work done on the two moving charged particles is

1 Same as the work done by charged particle on the stationary proton

2 Same as the same force law is involved in the experiments

3 Less for the case of a position, as the positron moves away more rapidly and the force on it weakness

4 More for the case of a positron, as the positron moves away a larger distance

Explanation:

Force between two protons = force between a proton and a positron. As positron is much lighter then proton, it moves away through much larger distance compared to proton. As work done $=$ force $\times$ distance, therefore in the same time $t$ , work done in a case of positron is more than that in case of proton.

NCERT Exemplar

PHXI06:WORK ENERGY AND POWER

355845 A body of mass $m$ is slowly pulled up the hill by a force $F$ which at each point was directed along the tangent of the trajectory as shown in figure. All surfaces are smooth. Find the work performed by this force.
supporting img

1

- m g l

2

m g l

3 Zero

4

m g h

Explanation:

$\sum W_{n e t} = Δ K$
As particle is moving slowly, this means
$\begin{aligned} Δ K = 0 \\ W_{N} + W_{F} + W_{m g} = Δ K \end{aligned}$
But $W_{N} = 0$ as $\vec{N} ⊥ \vec{d r}$
$\begin{aligned} 0 + W_{F} - m g h = 0 \\ W_{F} = m g h \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355846 The work done on a particle of mass $m$ by a force, $K [\frac{x}{{(x^{2} + y^{2})}^{3 / 2}} \hat{i} + \frac{y}{{(x^{2} + y^{2})}^{3 / 2}} \hat{j}]$ . being a constant of appropriate dimensions),
when the particle is taken from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius $a$ about the origin in the $x - y$ plane is

1 0

2

\frac{K π}{a}

3

\frac{2 K π}{a}

4

\frac{K π}{2 a}

Explanation:

Here $\frac{\partial F_{x}}{\partial y} = \frac{\partial F_{y}}{\partial x}$
That's why force is conservative and particle is moving on circle i.e., $x^{2} + y^{2} = a^{2}$
Here work is independent of path
Hence $W = \int_{a}^{0} F_{x} d x + F_{y} d y$
$= \frac{K}{a^{3}} \int_{a}^{0} x d x + \frac{K}{a^{3}} \int_{0}^{a} y d y = 0$

PHXI06:WORK ENERGY AND POWER

355843 A body of mass $m$ starts moving from rest along $x$ -axis so that its velocity varies as $v = a \sqrt{s}$ where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

1

\frac{1}{4} m a t^{2}

2

4 m a^{4} t^{2}

3

\frac{1}{8} m a^{4} t^{2}

4

8 m a^{4} t^{2}

Explanation:

$v = a \sqrt{s} = \frac{d s}{d t}$
$\int_{0}^{s} \frac{d s}{\sqrt{s}} = a \int_{0}^{t} d t$
$2 \sqrt{s} = a t$
$v = \frac{a^{2} t}{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{8} m a^{4} t^{2}$

JEE - 2018

PHXI06:WORK ENERGY AND POWER

355844 A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$ , the work done on the two moving charged particles is

1 Same as the work done by charged particle on the stationary proton

2 Same as the same force law is involved in the experiments

3 Less for the case of a position, as the positron moves away more rapidly and the force on it weakness

4 More for the case of a positron, as the positron moves away a larger distance

Explanation:

Force between two protons = force between a proton and a positron. As positron is much lighter then proton, it moves away through much larger distance compared to proton. As work done $=$ force $\times$ distance, therefore in the same time $t$ , work done in a case of positron is more than that in case of proton.

NCERT Exemplar

PHXI06:WORK ENERGY AND POWER

355845 A body of mass $m$ is slowly pulled up the hill by a force $F$ which at each point was directed along the tangent of the trajectory as shown in figure. All surfaces are smooth. Find the work performed by this force.
supporting img

1

- m g l

2

m g l

3 Zero

4

m g h

Explanation:

$\sum W_{n e t} = Δ K$
As particle is moving slowly, this means
$\begin{aligned} Δ K = 0 \\ W_{N} + W_{F} + W_{m g} = Δ K \end{aligned}$
But $W_{N} = 0$ as $\vec{N} ⊥ \vec{d r}$
$\begin{aligned} 0 + W_{F} - m g h = 0 \\ W_{F} = m g h \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355846 The work done on a particle of mass $m$ by a force, $K [\frac{x}{{(x^{2} + y^{2})}^{3 / 2}} \hat{i} + \frac{y}{{(x^{2} + y^{2})}^{3 / 2}} \hat{j}]$ . being a constant of appropriate dimensions),
when the particle is taken from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius $a$ about the origin in the $x - y$ plane is

1 0

2

\frac{K π}{a}

3

\frac{2 K π}{a}

4

\frac{K π}{2 a}

Explanation:

Here $\frac{\partial F_{x}}{\partial y} = \frac{\partial F_{y}}{\partial x}$
That's why force is conservative and particle is moving on circle i.e., $x^{2} + y^{2} = a^{2}$
Here work is independent of path
Hence $W = \int_{a}^{0} F_{x} d x + F_{y} d y$
$= \frac{K}{a^{3}} \int_{a}^{0} x d x + \frac{K}{a^{3}} \int_{0}^{a} y d y = 0$

PHXI06:WORK ENERGY AND POWER

355843 A body of mass $m$ starts moving from rest along $x$ -axis so that its velocity varies as $v = a \sqrt{s}$ where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

1

\frac{1}{4} m a t^{2}

2

4 m a^{4} t^{2}

3

\frac{1}{8} m a^{4} t^{2}

4

8 m a^{4} t^{2}

Explanation:

$v = a \sqrt{s} = \frac{d s}{d t}$
$\int_{0}^{s} \frac{d s}{\sqrt{s}} = a \int_{0}^{t} d t$
$2 \sqrt{s} = a t$
$v = \frac{a^{2} t}{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{8} m a^{4} t^{2}$

JEE - 2018

PHXI06:WORK ENERGY AND POWER

355844 A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$ , the work done on the two moving charged particles is

1 Same as the work done by charged particle on the stationary proton

2 Same as the same force law is involved in the experiments

3 Less for the case of a position, as the positron moves away more rapidly and the force on it weakness

4 More for the case of a positron, as the positron moves away a larger distance

Explanation:

Force between two protons = force between a proton and a positron. As positron is much lighter then proton, it moves away through much larger distance compared to proton. As work done $=$ force $\times$ distance, therefore in the same time $t$ , work done in a case of positron is more than that in case of proton.

NCERT Exemplar

PHXI06:WORK ENERGY AND POWER

355845 A body of mass $m$ is slowly pulled up the hill by a force $F$ which at each point was directed along the tangent of the trajectory as shown in figure. All surfaces are smooth. Find the work performed by this force.
supporting img

1

- m g l

2

m g l

3 Zero

4

m g h

Explanation:

$\sum W_{n e t} = Δ K$
As particle is moving slowly, this means
$\begin{aligned} Δ K = 0 \\ W_{N} + W_{F} + W_{m g} = Δ K \end{aligned}$
But $W_{N} = 0$ as $\vec{N} ⊥ \vec{d r}$
$\begin{aligned} 0 + W_{F} - m g h = 0 \\ W_{F} = m g h \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355846 The work done on a particle of mass $m$ by a force, $K [\frac{x}{{(x^{2} + y^{2})}^{3 / 2}} \hat{i} + \frac{y}{{(x^{2} + y^{2})}^{3 / 2}} \hat{j}]$ . being a constant of appropriate dimensions),
when the particle is taken from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius $a$ about the origin in the $x - y$ plane is

1 0

2

\frac{K π}{a}

3

\frac{2 K π}{a}

4

\frac{K π}{2 a}

Explanation:

Here $\frac{\partial F_{x}}{\partial y} = \frac{\partial F_{y}}{\partial x}$
That's why force is conservative and particle is moving on circle i.e., $x^{2} + y^{2} = a^{2}$
Here work is independent of path
Hence $W = \int_{a}^{0} F_{x} d x + F_{y} d y$
$= \frac{K}{a^{3}} \int_{a}^{0} x d x + \frac{K}{a^{3}} \int_{0}^{a} y d y = 0$

PHXI06:WORK ENERGY AND POWER

355843 A body of mass $m$ starts moving from rest along $x$ -axis so that its velocity varies as $v = a \sqrt{s}$ where $a$ is a constant and $s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first $t$ seconds after the start of the motion is:

1

\frac{1}{4} m a t^{2}

2

4 m a^{4} t^{2}

3

\frac{1}{8} m a^{4} t^{2}

4

8 m a^{4} t^{2}

Explanation:

$v = a \sqrt{s} = \frac{d s}{d t}$
$\int_{0}^{s} \frac{d s}{\sqrt{s}} = a \int_{0}^{t} d t$
$2 \sqrt{s} = a t$
$v = \frac{a^{2} t}{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{8} m a^{4} t^{2}$

JEE - 2018

PHXI06:WORK ENERGY AND POWER

355844 A proton is kept at rest. A positively charged particle is released from rest at a distance $d$ in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time $t$ , the work done on the two moving charged particles is

1 Same as the work done by charged particle on the stationary proton

2 Same as the same force law is involved in the experiments

3 Less for the case of a position, as the positron moves away more rapidly and the force on it weakness

4 More for the case of a positron, as the positron moves away a larger distance

Explanation:

Force between two protons = force between a proton and a positron. As positron is much lighter then proton, it moves away through much larger distance compared to proton. As work done $=$ force $\times$ distance, therefore in the same time $t$ , work done in a case of positron is more than that in case of proton.

NCERT Exemplar

PHXI06:WORK ENERGY AND POWER

355845 A body of mass $m$ is slowly pulled up the hill by a force $F$ which at each point was directed along the tangent of the trajectory as shown in figure. All surfaces are smooth. Find the work performed by this force.
supporting img

1

- m g l

2

m g l

3 Zero

4

m g h

Explanation:

$\sum W_{n e t} = Δ K$
As particle is moving slowly, this means
$\begin{aligned} Δ K = 0 \\ W_{N} + W_{F} + W_{m g} = Δ K \end{aligned}$
But $W_{N} = 0$ as $\vec{N} ⊥ \vec{d r}$
$\begin{aligned} 0 + W_{F} - m g h = 0 \\ W_{F} = m g h \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355846 The work done on a particle of mass $m$ by a force, $K [\frac{x}{{(x^{2} + y^{2})}^{3 / 2}} \hat{i} + \frac{y}{{(x^{2} + y^{2})}^{3 / 2}} \hat{j}]$ . being a constant of appropriate dimensions),
when the particle is taken from the point $(a, 0)$ to the point $(0, a)$ along a circular path of radius $a$ about the origin in the $x - y$ plane is

1 0

2

\frac{K π}{a}

3

\frac{2 K π}{a}

4

\frac{K π}{2 a}

Explanation:

Here $\frac{\partial F_{x}}{\partial y} = \frac{\partial F_{y}}{\partial x}$
That's why force is conservative and particle is moving on circle i.e., $x^{2} + y^{2} = a^{2}$
Here work is independent of path
Hence $W = \int_{a}^{0} F_{x} d x + F_{y} d y$
$= \frac{K}{a^{3}} \int_{a}^{0} x d x + \frac{K}{a^{3}} \int_{0}^{a} y d y = 0$