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PHXI06:WORK ENERGY AND POWER

355625 A ball is thrown vertically upwards with a velocity of 10 $m / s$ . It returns to the ground with a velocity of 9 $m / s$ . If $g = 9.8 m / s^{2}$ , then the maximum height attained by the is nearly ( assume air resistance to be uniform)

1 5.1

m

2 4.1

m

3 4.61

m

4 5.0

m

Explanation:

Total loss in friction $= K_{i} - K_{f}$
$= \frac{1}{2} m (100 - 81)$
Half of the loss will be in upward direction Hence,
$m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
Therefore, $h = 4.61 m$

PHXI06:WORK ENERGY AND POWER

355626 Given that the displacement of the body in metre is a function of time as follows $x = 2 t^{4} + 5$ . The mass of the body is 2 $k g$ . What is the increase in its kinetic energy one second after that start of motion?

1 8

J

2 16

J

3 32

J

4 64

J

Explanation:

$v = \frac{d x}{d t} = 8 t^{3}$
$v_{0 s} = 0, v_{1 s} = 8 m / s$
$∴ Δ K E = \frac{1}{2} \times 2 \times (64 - 0) J$
$= 64 J$

PHXI06:WORK ENERGY AND POWER

355627 Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P = 6 m$ and $O Q = 8 m$ . A block of mass $1 k g$ is pulled along the rail from $P$ to $Q$ with a force of $15 N,$ which is always parallel to line $P Q$ (see the figure given).
Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules.
The value of $n$ is
(Take acceleration due to gravity $= 10 m s^{- 2}$ )
supporting img

1 3

2 7

3 9

4 12

Explanation:

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$

PHXI06:WORK ENERGY AND POWER

355628 A particle moves in a straight line with retardation proportional to its displacement. The loss in kinetic energy of the particle, for any displacement $x$ is proportional to:

1

x^{2}

2

e^{x}

3

x

4

\log_{e}^{x}

Explanation:

$a = - k x$ or $\frac{v d v}{d x} = - k x, v d v = - k x d x$
Let the velocity change from $v_{0}$ to $v$
$\int_{v_{0}}^{v} v d v = - \int_{0}^{x} k x d x \Rightarrow \frac{v^{2} - v_{0}^{2}}{2} = \frac{- k x^{2}}{2}$
$m (\frac{v^{2} - v_{0}^{2}}{2}) = \frac{- m k x^{2}}{2}$
$Δ k \propto x^{2} [Δ k$ is loss in kinetic energy]

PHXI06:WORK ENERGY AND POWER

355629 A block of mass 10 $k g$ is moving in $x$ - direction with a constant speed of 10 $m / s$ . It is subjected to a retarding force $F = (0.1 x) N$ during its travel from $x = 20 m$ to $x = 30 m$ . Its final kinetic energy will be

1 475

J

2 450

J

3 275

J

4 250

J

Explanation:

$K_{f} - K_{i} = W$
$∴ K_{f} = - K_{i} + W$
$= \frac{1}{2} \times 10 \times {(10)}^{2} + \int_{20}^{30} (- 0.1 x) d x$
$= 475 J$

PHXI06:WORK ENERGY AND POWER

355625 A ball is thrown vertically upwards with a velocity of 10 $m / s$ . It returns to the ground with a velocity of 9 $m / s$ . If $g = 9.8 m / s^{2}$ , then the maximum height attained by the is nearly ( assume air resistance to be uniform)

1 5.1

m

2 4.1

m

3 4.61

m

4 5.0

m

Explanation:

Total loss in friction $= K_{i} - K_{f}$
$= \frac{1}{2} m (100 - 81)$
Half of the loss will be in upward direction Hence,
$m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
Therefore, $h = 4.61 m$

PHXI06:WORK ENERGY AND POWER

355626 Given that the displacement of the body in metre is a function of time as follows $x = 2 t^{4} + 5$ . The mass of the body is 2 $k g$ . What is the increase in its kinetic energy one second after that start of motion?

1 8

J

2 16

J

3 32

J

4 64

J

Explanation:

$v = \frac{d x}{d t} = 8 t^{3}$
$v_{0 s} = 0, v_{1 s} = 8 m / s$
$∴ Δ K E = \frac{1}{2} \times 2 \times (64 - 0) J$
$= 64 J$

PHXI06:WORK ENERGY AND POWER

355627 Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P = 6 m$ and $O Q = 8 m$ . A block of mass $1 k g$ is pulled along the rail from $P$ to $Q$ with a force of $15 N,$ which is always parallel to line $P Q$ (see the figure given).
Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules.
The value of $n$ is
(Take acceleration due to gravity $= 10 m s^{- 2}$ )
supporting img

1 3

2 7

3 9

4 12

Explanation:

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$

PHXI06:WORK ENERGY AND POWER

355628 A particle moves in a straight line with retardation proportional to its displacement. The loss in kinetic energy of the particle, for any displacement $x$ is proportional to:

1

x^{2}

2

e^{x}

3

x

4

\log_{e}^{x}

Explanation:

$a = - k x$ or $\frac{v d v}{d x} = - k x, v d v = - k x d x$
Let the velocity change from $v_{0}$ to $v$
$\int_{v_{0}}^{v} v d v = - \int_{0}^{x} k x d x \Rightarrow \frac{v^{2} - v_{0}^{2}}{2} = \frac{- k x^{2}}{2}$
$m (\frac{v^{2} - v_{0}^{2}}{2}) = \frac{- m k x^{2}}{2}$
$Δ k \propto x^{2} [Δ k$ is loss in kinetic energy]

PHXI06:WORK ENERGY AND POWER

355629 A block of mass 10 $k g$ is moving in $x$ - direction with a constant speed of 10 $m / s$ . It is subjected to a retarding force $F = (0.1 x) N$ during its travel from $x = 20 m$ to $x = 30 m$ . Its final kinetic energy will be

1 475

J

2 450

J

3 275

J

4 250

J

Explanation:

$K_{f} - K_{i} = W$
$∴ K_{f} = - K_{i} + W$
$= \frac{1}{2} \times 10 \times {(10)}^{2} + \int_{20}^{30} (- 0.1 x) d x$
$= 475 J$

PHXI06:WORK ENERGY AND POWER

355625 A ball is thrown vertically upwards with a velocity of 10 $m / s$ . It returns to the ground with a velocity of 9 $m / s$ . If $g = 9.8 m / s^{2}$ , then the maximum height attained by the is nearly ( assume air resistance to be uniform)

1 5.1

m

2 4.1

m

3 4.61

m

4 5.0

m

Explanation:

Total loss in friction $= K_{i} - K_{f}$
$= \frac{1}{2} m (100 - 81)$
Half of the loss will be in upward direction Hence,
$m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
Therefore, $h = 4.61 m$

PHXI06:WORK ENERGY AND POWER

355626 Given that the displacement of the body in metre is a function of time as follows $x = 2 t^{4} + 5$ . The mass of the body is 2 $k g$ . What is the increase in its kinetic energy one second after that start of motion?

1 8

J

2 16

J

3 32

J

4 64

J

Explanation:

$v = \frac{d x}{d t} = 8 t^{3}$
$v_{0 s} = 0, v_{1 s} = 8 m / s$
$∴ Δ K E = \frac{1}{2} \times 2 \times (64 - 0) J$
$= 64 J$

PHXI06:WORK ENERGY AND POWER

355627 Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P = 6 m$ and $O Q = 8 m$ . A block of mass $1 k g$ is pulled along the rail from $P$ to $Q$ with a force of $15 N,$ which is always parallel to line $P Q$ (see the figure given).
Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules.
The value of $n$ is
(Take acceleration due to gravity $= 10 m s^{- 2}$ )
supporting img

1 3

2 7

3 9

4 12

Explanation:

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$

PHXI06:WORK ENERGY AND POWER

355628 A particle moves in a straight line with retardation proportional to its displacement. The loss in kinetic energy of the particle, for any displacement $x$ is proportional to:

1

x^{2}

2

e^{x}

3

x

4

\log_{e}^{x}

Explanation:

$a = - k x$ or $\frac{v d v}{d x} = - k x, v d v = - k x d x$
Let the velocity change from $v_{0}$ to $v$
$\int_{v_{0}}^{v} v d v = - \int_{0}^{x} k x d x \Rightarrow \frac{v^{2} - v_{0}^{2}}{2} = \frac{- k x^{2}}{2}$
$m (\frac{v^{2} - v_{0}^{2}}{2}) = \frac{- m k x^{2}}{2}$
$Δ k \propto x^{2} [Δ k$ is loss in kinetic energy]

PHXI06:WORK ENERGY AND POWER

355629 A block of mass 10 $k g$ is moving in $x$ - direction with a constant speed of 10 $m / s$ . It is subjected to a retarding force $F = (0.1 x) N$ during its travel from $x = 20 m$ to $x = 30 m$ . Its final kinetic energy will be

1 475

J

2 450

J

3 275

J

4 250

J

Explanation:

$K_{f} - K_{i} = W$
$∴ K_{f} = - K_{i} + W$
$= \frac{1}{2} \times 10 \times {(10)}^{2} + \int_{20}^{30} (- 0.1 x) d x$
$= 475 J$

PHXI06:WORK ENERGY AND POWER

355625 A ball is thrown vertically upwards with a velocity of 10 $m / s$ . It returns to the ground with a velocity of 9 $m / s$ . If $g = 9.8 m / s^{2}$ , then the maximum height attained by the is nearly ( assume air resistance to be uniform)

1 5.1

m

2 4.1

m

3 4.61

m

4 5.0

m

Explanation:

Total loss in friction $= K_{i} - K_{f}$
$= \frac{1}{2} m (100 - 81)$
Half of the loss will be in upward direction Hence,
$m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
Therefore, $h = 4.61 m$

PHXI06:WORK ENERGY AND POWER

355626 Given that the displacement of the body in metre is a function of time as follows $x = 2 t^{4} + 5$ . The mass of the body is 2 $k g$ . What is the increase in its kinetic energy one second after that start of motion?

1 8

J

2 16

J

3 32

J

4 64

J

Explanation:

$v = \frac{d x}{d t} = 8 t^{3}$
$v_{0 s} = 0, v_{1 s} = 8 m / s$
$∴ Δ K E = \frac{1}{2} \times 2 \times (64 - 0) J$
$= 64 J$

PHXI06:WORK ENERGY AND POWER

355627 Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P = 6 m$ and $O Q = 8 m$ . A block of mass $1 k g$ is pulled along the rail from $P$ to $Q$ with a force of $15 N,$ which is always parallel to line $P Q$ (see the figure given).
Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules.
The value of $n$ is
(Take acceleration due to gravity $= 10 m s^{- 2}$ )
supporting img

1 3

2 7

3 9

4 12

Explanation:

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$

PHXI06:WORK ENERGY AND POWER

355628 A particle moves in a straight line with retardation proportional to its displacement. The loss in kinetic energy of the particle, for any displacement $x$ is proportional to:

1

x^{2}

2

e^{x}

3

x

4

\log_{e}^{x}

Explanation:

$a = - k x$ or $\frac{v d v}{d x} = - k x, v d v = - k x d x$
Let the velocity change from $v_{0}$ to $v$
$\int_{v_{0}}^{v} v d v = - \int_{0}^{x} k x d x \Rightarrow \frac{v^{2} - v_{0}^{2}}{2} = \frac{- k x^{2}}{2}$
$m (\frac{v^{2} - v_{0}^{2}}{2}) = \frac{- m k x^{2}}{2}$
$Δ k \propto x^{2} [Δ k$ is loss in kinetic energy]

PHXI06:WORK ENERGY AND POWER

355629 A block of mass 10 $k g$ is moving in $x$ - direction with a constant speed of 10 $m / s$ . It is subjected to a retarding force $F = (0.1 x) N$ during its travel from $x = 20 m$ to $x = 30 m$ . Its final kinetic energy will be

1 475

J

2 450

J

3 275

J

4 250

J

Explanation:

$K_{f} - K_{i} = W$
$∴ K_{f} = - K_{i} + W$
$= \frac{1}{2} \times 10 \times {(10)}^{2} + \int_{20}^{30} (- 0.1 x) d x$
$= 475 J$

PHXI06:WORK ENERGY AND POWER

355625 A ball is thrown vertically upwards with a velocity of 10 $m / s$ . It returns to the ground with a velocity of 9 $m / s$ . If $g = 9.8 m / s^{2}$ , then the maximum height attained by the is nearly ( assume air resistance to be uniform)

1 5.1

m

2 4.1

m

3 4.61

m

4 5.0

m

Explanation:

Total loss in friction $= K_{i} - K_{f}$
$= \frac{1}{2} m (100 - 81)$
Half of the loss will be in upward direction Hence,
$m g h + \frac{1}{2} (\frac{19}{2} m) = K_{i} = \frac{1}{2} m (100) = 50 m$
Therefore, $h = 4.61 m$

PHXI06:WORK ENERGY AND POWER

355626 Given that the displacement of the body in metre is a function of time as follows $x = 2 t^{4} + 5$ . The mass of the body is 2 $k g$ . What is the increase in its kinetic energy one second after that start of motion?

1 8

J

2 16

J

3 32

J

4 64

J

Explanation:

$v = \frac{d x}{d t} = 8 t^{3}$
$v_{0 s} = 0, v_{1 s} = 8 m / s$
$∴ Δ K E = \frac{1}{2} \times 2 \times (64 - 0) J$
$= 64 J$

PHXI06:WORK ENERGY AND POWER

355627 Consider an elliptically shaped rail $P Q$ in the vertical plane with $O P = 6 m$ and $O Q = 8 m$ . A block of mass $1 k g$ is pulled along the rail from $P$ to $Q$ with a force of $15 N,$ which is always parallel to line $P Q$ (see the figure given).
Assuming no frictional losses, the kinetic energy of the block when it reaches $Q$ is $(n \times 10)$ joules.
The value of $n$ is
(Take acceleration due to gravity $= 10 m s^{- 2}$ )
supporting img

1 3

2 7

3 9

4 12

Explanation:

Using work energy theorem
$W_{m g} + W_{F} = Δ K . E .$
$- m g h + F d = Δ K . E$
$∴ - 1 \times 10 \times 8 + 15 (10) = Δ K E$
$∴ Δ K . E . = 70$
$\Rightarrow n = 7$

PHXI06:WORK ENERGY AND POWER

355628 A particle moves in a straight line with retardation proportional to its displacement. The loss in kinetic energy of the particle, for any displacement $x$ is proportional to:

1

x^{2}

2

e^{x}

3

x

4

\log_{e}^{x}

Explanation:

$a = - k x$ or $\frac{v d v}{d x} = - k x, v d v = - k x d x$
Let the velocity change from $v_{0}$ to $v$
$\int_{v_{0}}^{v} v d v = - \int_{0}^{x} k x d x \Rightarrow \frac{v^{2} - v_{0}^{2}}{2} = \frac{- k x^{2}}{2}$
$m (\frac{v^{2} - v_{0}^{2}}{2}) = \frac{- m k x^{2}}{2}$
$Δ k \propto x^{2} [Δ k$ is loss in kinetic energy]

PHXI06:WORK ENERGY AND POWER

355629 A block of mass 10 $k g$ is moving in $x$ - direction with a constant speed of 10 $m / s$ . It is subjected to a retarding force $F = (0.1 x) N$ during its travel from $x = 20 m$ to $x = 30 m$ . Its final kinetic energy will be

1 475

J

2 450

J

3 275

J

4 250

J

Explanation:

$K_{f} - K_{i} = W$
$∴ K_{f} = - K_{i} + W$
$= \frac{1}{2} \times 10 \times {(10)}^{2} + \int_{20}^{30} (- 0.1 x) d x$
$= 475 J$