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PHXI06:WORK ENERGY AND POWER

355847 A particle is moved from $(0, 0)$ to $(2 m, 2 m)$ with force $\vec{F} = (x \hat{j}) N$ along path (i) $y = x$ (ii) $y = \sqrt{x}$ . The work done about two paths is:

1

2 J, 2 J

2

2 J, \frac{8}{3} J

3

\frac{8}{3} J, \frac{8}{3} J

4

9 J, 8 J

Explanation:

$d W = \vec{F} \cdot d \vec{s} = (x \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = x d y$
(i) when $y = x$
$W = \int d W = \int_{0}^{2} y d y = {| \frac{y^{2}}{2} |}_{0}^{2} = 2 J$
(ii) When $y = \sqrt{x}$ or $y^{2} = x$
$W = \int_{0}^{2} y^{2} d y = {| \frac{y^{3}}{3} |}_{0}^{2} = \frac{8}{3} J$

PHXI06:WORK ENERGY AND POWER

355848 A chain of linear mass density 3 $k g / m$ and length $8 m$ is lying on the table with 4 $m$ of chain hanging from the edge. The work done in lifting the chain on the table will be -

1 235.2

J

2 117.6

J

3 196

J

4 98

J

Explanation:

Mass of the chain hanging $= 4 \times 3 = 12 kg$
Shift in centre of gravity $= 4 / 2 = 2 m$
$W = m g h = 12 \times 9.8 \times 2 = 235.2 J$

PHXI06:WORK ENERGY AND POWER

355849 Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = K v^{- 2}$ , where $K$ is constant. The work done by this force in time $t$ is:

1

K v t

2

K^{2} v^{2} t^{2}

3

K^{2} v t

4

\frac{3 K t}{2 v}

Explanation:

$m \frac{d v}{d t} = K v^{- 2} \int m (v^{2} d v) = \int K d t$
$m (\frac{v^{3}}{3}) = K t \frac{1}{2} m v^{2} = \frac{3}{2} \frac{K t}{v}$

PHXI06:WORK ENERGY AND POWER

355850 A chain of mass $M$ is placed on a smooth table with $1 / n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is

1

M g L / n

2

M g L / 2 n

3

M g L / n^{2}

4

M g L / 2 n^{2}

Explanation:

$W =$ change in $P E$ of $C O M$ of hanging part
$= \frac{M}{n} g \frac{L}{2 n} = \frac{M g L}{2 n^{2}}$

PHXI06:WORK ENERGY AND POWER

355847 A particle is moved from $(0, 0)$ to $(2 m, 2 m)$ with force $\vec{F} = (x \hat{j}) N$ along path (i) $y = x$ (ii) $y = \sqrt{x}$ . The work done about two paths is:

1

2 J, 2 J

2

2 J, \frac{8}{3} J

3

\frac{8}{3} J, \frac{8}{3} J

4

9 J, 8 J

Explanation:

$d W = \vec{F} \cdot d \vec{s} = (x \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = x d y$
(i) when $y = x$
$W = \int d W = \int_{0}^{2} y d y = {| \frac{y^{2}}{2} |}_{0}^{2} = 2 J$
(ii) When $y = \sqrt{x}$ or $y^{2} = x$
$W = \int_{0}^{2} y^{2} d y = {| \frac{y^{3}}{3} |}_{0}^{2} = \frac{8}{3} J$

PHXI06:WORK ENERGY AND POWER

355848 A chain of linear mass density 3 $k g / m$ and length $8 m$ is lying on the table with 4 $m$ of chain hanging from the edge. The work done in lifting the chain on the table will be -

1 235.2

J

2 117.6

J

3 196

J

4 98

J

Explanation:

Mass of the chain hanging $= 4 \times 3 = 12 kg$
Shift in centre of gravity $= 4 / 2 = 2 m$
$W = m g h = 12 \times 9.8 \times 2 = 235.2 J$

PHXI06:WORK ENERGY AND POWER

355849 Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = K v^{- 2}$ , where $K$ is constant. The work done by this force in time $t$ is:

1

K v t

2

K^{2} v^{2} t^{2}

3

K^{2} v t

4

\frac{3 K t}{2 v}

Explanation:

$m \frac{d v}{d t} = K v^{- 2} \int m (v^{2} d v) = \int K d t$
$m (\frac{v^{3}}{3}) = K t \frac{1}{2} m v^{2} = \frac{3}{2} \frac{K t}{v}$

PHXI06:WORK ENERGY AND POWER

355850 A chain of mass $M$ is placed on a smooth table with $1 / n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is

1

M g L / n

2

M g L / 2 n

3

M g L / n^{2}

4

M g L / 2 n^{2}

Explanation:

$W =$ change in $P E$ of $C O M$ of hanging part
$= \frac{M}{n} g \frac{L}{2 n} = \frac{M g L}{2 n^{2}}$

PHXI06:WORK ENERGY AND POWER

355847 A particle is moved from $(0, 0)$ to $(2 m, 2 m)$ with force $\vec{F} = (x \hat{j}) N$ along path (i) $y = x$ (ii) $y = \sqrt{x}$ . The work done about two paths is:

1

2 J, 2 J

2

2 J, \frac{8}{3} J

3

\frac{8}{3} J, \frac{8}{3} J

4

9 J, 8 J

Explanation:

$d W = \vec{F} \cdot d \vec{s} = (x \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = x d y$
(i) when $y = x$
$W = \int d W = \int_{0}^{2} y d y = {| \frac{y^{2}}{2} |}_{0}^{2} = 2 J$
(ii) When $y = \sqrt{x}$ or $y^{2} = x$
$W = \int_{0}^{2} y^{2} d y = {| \frac{y^{3}}{3} |}_{0}^{2} = \frac{8}{3} J$

PHXI06:WORK ENERGY AND POWER

355848 A chain of linear mass density 3 $k g / m$ and length $8 m$ is lying on the table with 4 $m$ of chain hanging from the edge. The work done in lifting the chain on the table will be -

1 235.2

J

2 117.6

J

3 196

J

4 98

J

Explanation:

Mass of the chain hanging $= 4 \times 3 = 12 kg$
Shift in centre of gravity $= 4 / 2 = 2 m$
$W = m g h = 12 \times 9.8 \times 2 = 235.2 J$

PHXI06:WORK ENERGY AND POWER

355849 Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = K v^{- 2}$ , where $K$ is constant. The work done by this force in time $t$ is:

1

K v t

2

K^{2} v^{2} t^{2}

3

K^{2} v t

4

\frac{3 K t}{2 v}

Explanation:

$m \frac{d v}{d t} = K v^{- 2} \int m (v^{2} d v) = \int K d t$
$m (\frac{v^{3}}{3}) = K t \frac{1}{2} m v^{2} = \frac{3}{2} \frac{K t}{v}$

PHXI06:WORK ENERGY AND POWER

355850 A chain of mass $M$ is placed on a smooth table with $1 / n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is

1

M g L / n

2

M g L / 2 n

3

M g L / n^{2}

4

M g L / 2 n^{2}

Explanation:

$W =$ change in $P E$ of $C O M$ of hanging part
$= \frac{M}{n} g \frac{L}{2 n} = \frac{M g L}{2 n^{2}}$

PHXI06:WORK ENERGY AND POWER

355847 A particle is moved from $(0, 0)$ to $(2 m, 2 m)$ with force $\vec{F} = (x \hat{j}) N$ along path (i) $y = x$ (ii) $y = \sqrt{x}$ . The work done about two paths is:

1

2 J, 2 J

2

2 J, \frac{8}{3} J

3

\frac{8}{3} J, \frac{8}{3} J

4

9 J, 8 J

Explanation:

$d W = \vec{F} \cdot d \vec{s} = (x \hat{j}) \cdot (d x \hat{i} + d y \hat{j}) = x d y$
(i) when $y = x$
$W = \int d W = \int_{0}^{2} y d y = {| \frac{y^{2}}{2} |}_{0}^{2} = 2 J$
(ii) When $y = \sqrt{x}$ or $y^{2} = x$
$W = \int_{0}^{2} y^{2} d y = {| \frac{y^{3}}{3} |}_{0}^{2} = \frac{8}{3} J$

PHXI06:WORK ENERGY AND POWER

355848 A chain of linear mass density 3 $k g / m$ and length $8 m$ is lying on the table with 4 $m$ of chain hanging from the edge. The work done in lifting the chain on the table will be -

1 235.2

J

2 117.6

J

3 196

J

4 98

J

Explanation:

Mass of the chain hanging $= 4 \times 3 = 12 kg$
Shift in centre of gravity $= 4 / 2 = 2 m$
$W = m g h = 12 \times 9.8 \times 2 = 235.2 J$

PHXI06:WORK ENERGY AND POWER

355849 Force acting on a particle moving in a straight line varies with the velocity of the particle as $F = K v^{- 2}$ , where $K$ is constant. The work done by this force in time $t$ is:

1

K v t

2

K^{2} v^{2} t^{2}

3

K^{2} v t

4

\frac{3 K t}{2 v}

Explanation:

$m \frac{d v}{d t} = K v^{- 2} \int m (v^{2} d v) = \int K d t$
$m (\frac{v^{3}}{3}) = K t \frac{1}{2} m v^{2} = \frac{3}{2} \frac{K t}{v}$

PHXI06:WORK ENERGY AND POWER

355850 A chain of mass $M$ is placed on a smooth table with $1 / n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is

1

M g L / n

2

M g L / 2 n

3

M g L / n^{2}

4

M g L / 2 n^{2}

Explanation:

$W =$ change in $P E$ of $C O M$ of hanging part
$= \frac{M}{n} g \frac{L}{2 n} = \frac{M g L}{2 n^{2}}$

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