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PHXI06:WORK ENERGY AND POWER

355611 A mass of 0.5 $k g$ moving with a speed of 15 $m / s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k = 50 N / m$ . The maximum compression of the spring would be:
supporting img

1 0.15

m

2 0.5

m

3 1.5

m

4 0.12

m

Explanation:

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\begin{aligned} \Rightarrow m v^{2} = k x^{2} \Rightarrow 0.5 \times (1.5)^{2} = 50 \times x^{2} \\ ∴ x = 0.15 m \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355612 The spring extends by $x$ on loading, then energy stored by the spring is : (if $T$ is the tension in spring and $k$ is spring constant)

1

\frac{2 T^{2}}{k}

2

\frac{T^{2}}{2 k^{2}}

3

\frac{T^{2}}{2 k}

4

\frac{2 k}{T^{2}}

Explanation:

$U = \frac{F^{2}}{2 k} = \frac{T^{2}}{2 k}$ .

PHXI06:WORK ENERGY AND POWER

355613 An elastic string of unstretched length $L$ and force constant $k$ is stretched by another small $x$ . It is further stretched by another small length $y$ . The work done in the second stretching is

1

\frac{1}{2} k y (2 x + y)

2

\frac{1}{2} k y^{2}

3

\frac{1}{2} k (x^{2} + y^{2})

4

\frac{1}{2} k (x + y)^{2}

Explanation:

$W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x + y)^{2}$ $W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x^{2} + y^{2} + 2 x y)$
$= \frac{1}{2} k y (- 2 x - y)$
The work done against elastic force is
$W_{e x t} = - W = \frac{k y}{2} (2 x + y)$

PHXI06:WORK ENERGY AND POWER

355614 A string of length $L$ and force constant $K$ is stretched to obtain extension $l$ . It is further stretched to obtain extension $l_{1}$ . The work done in second stretching is

1

\frac{1}{2} K l_{1}^{2}

2

\frac{1}{2} K (l_{1}^{2} - l^{2})

3

\frac{1}{2} K (l_{1}^{2} + l^{2})

4

\frac{1}{2} K l_{1} (2 l + l_{1})

Explanation:

Work done in stretching a string to obtain an extension $l$ is
$W_{1} = \frac{1}{2} K l$
Similarly, work done in stretching a string to obtain extension $l_{1}$ is
$W_{2} = \frac{1}{2} K l_{1}^{2}$
$∴$ Work done in second case
$= W_{2} - W_{1} = \frac{1}{2} K (l_{1}^{2} - l^{2})$

PHXI06:WORK ENERGY AND POWER

355611 A mass of 0.5 $k g$ moving with a speed of 15 $m / s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k = 50 N / m$ . The maximum compression of the spring would be:
supporting img

1 0.15

m

2 0.5

m

3 1.5

m

4 0.12

m

Explanation:

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\begin{aligned} \Rightarrow m v^{2} = k x^{2} \Rightarrow 0.5 \times (1.5)^{2} = 50 \times x^{2} \\ ∴ x = 0.15 m \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355612 The spring extends by $x$ on loading, then energy stored by the spring is : (if $T$ is the tension in spring and $k$ is spring constant)

1

\frac{2 T^{2}}{k}

2

\frac{T^{2}}{2 k^{2}}

3

\frac{T^{2}}{2 k}

4

\frac{2 k}{T^{2}}

Explanation:

$U = \frac{F^{2}}{2 k} = \frac{T^{2}}{2 k}$ .

PHXI06:WORK ENERGY AND POWER

355613 An elastic string of unstretched length $L$ and force constant $k$ is stretched by another small $x$ . It is further stretched by another small length $y$ . The work done in the second stretching is

1

\frac{1}{2} k y (2 x + y)

2

\frac{1}{2} k y^{2}

3

\frac{1}{2} k (x^{2} + y^{2})

4

\frac{1}{2} k (x + y)^{2}

Explanation:

$W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x + y)^{2}$ $W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x^{2} + y^{2} + 2 x y)$
$= \frac{1}{2} k y (- 2 x - y)$
The work done against elastic force is
$W_{e x t} = - W = \frac{k y}{2} (2 x + y)$

PHXI06:WORK ENERGY AND POWER

355614 A string of length $L$ and force constant $K$ is stretched to obtain extension $l$ . It is further stretched to obtain extension $l_{1}$ . The work done in second stretching is

1

\frac{1}{2} K l_{1}^{2}

2

\frac{1}{2} K (l_{1}^{2} - l^{2})

3

\frac{1}{2} K (l_{1}^{2} + l^{2})

4

\frac{1}{2} K l_{1} (2 l + l_{1})

Explanation:

Work done in stretching a string to obtain an extension $l$ is
$W_{1} = \frac{1}{2} K l$
Similarly, work done in stretching a string to obtain extension $l_{1}$ is
$W_{2} = \frac{1}{2} K l_{1}^{2}$
$∴$ Work done in second case
$= W_{2} - W_{1} = \frac{1}{2} K (l_{1}^{2} - l^{2})$

PHXI06:WORK ENERGY AND POWER

355611 A mass of 0.5 $k g$ moving with a speed of 15 $m / s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k = 50 N / m$ . The maximum compression of the spring would be:
supporting img

1 0.15

m

2 0.5

m

3 1.5

m

4 0.12

m

Explanation:

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\begin{aligned} \Rightarrow m v^{2} = k x^{2} \Rightarrow 0.5 \times (1.5)^{2} = 50 \times x^{2} \\ ∴ x = 0.15 m \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355612 The spring extends by $x$ on loading, then energy stored by the spring is : (if $T$ is the tension in spring and $k$ is spring constant)

1

\frac{2 T^{2}}{k}

2

\frac{T^{2}}{2 k^{2}}

3

\frac{T^{2}}{2 k}

4

\frac{2 k}{T^{2}}

Explanation:

$U = \frac{F^{2}}{2 k} = \frac{T^{2}}{2 k}$ .

PHXI06:WORK ENERGY AND POWER

355613 An elastic string of unstretched length $L$ and force constant $k$ is stretched by another small $x$ . It is further stretched by another small length $y$ . The work done in the second stretching is

1

\frac{1}{2} k y (2 x + y)

2

\frac{1}{2} k y^{2}

3

\frac{1}{2} k (x^{2} + y^{2})

4

\frac{1}{2} k (x + y)^{2}

Explanation:

$W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x + y)^{2}$ $W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x^{2} + y^{2} + 2 x y)$
$= \frac{1}{2} k y (- 2 x - y)$
The work done against elastic force is
$W_{e x t} = - W = \frac{k y}{2} (2 x + y)$

PHXI06:WORK ENERGY AND POWER

355614 A string of length $L$ and force constant $K$ is stretched to obtain extension $l$ . It is further stretched to obtain extension $l_{1}$ . The work done in second stretching is

1

\frac{1}{2} K l_{1}^{2}

2

\frac{1}{2} K (l_{1}^{2} - l^{2})

3

\frac{1}{2} K (l_{1}^{2} + l^{2})

4

\frac{1}{2} K l_{1} (2 l + l_{1})

Explanation:

Work done in stretching a string to obtain an extension $l$ is
$W_{1} = \frac{1}{2} K l$
Similarly, work done in stretching a string to obtain extension $l_{1}$ is
$W_{2} = \frac{1}{2} K l_{1}^{2}$
$∴$ Work done in second case
$= W_{2} - W_{1} = \frac{1}{2} K (l_{1}^{2} - l^{2})$

PHXI06:WORK ENERGY AND POWER

355611 A mass of 0.5 $k g$ moving with a speed of 15 $m / s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k = 50 N / m$ . The maximum compression of the spring would be:
supporting img

1 0.15

m

2 0.5

m

3 1.5

m

4 0.12

m

Explanation:

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$
$\begin{aligned} \Rightarrow m v^{2} = k x^{2} \Rightarrow 0.5 \times (1.5)^{2} = 50 \times x^{2} \\ ∴ x = 0.15 m \end{aligned}$

PHXI06:WORK ENERGY AND POWER

355612 The spring extends by $x$ on loading, then energy stored by the spring is : (if $T$ is the tension in spring and $k$ is spring constant)

1

\frac{2 T^{2}}{k}

2

\frac{T^{2}}{2 k^{2}}

3

\frac{T^{2}}{2 k}

4

\frac{2 k}{T^{2}}

Explanation:

$U = \frac{F^{2}}{2 k} = \frac{T^{2}}{2 k}$ .

PHXI06:WORK ENERGY AND POWER

355613 An elastic string of unstretched length $L$ and force constant $k$ is stretched by another small $x$ . It is further stretched by another small length $y$ . The work done in the second stretching is

1

\frac{1}{2} k y (2 x + y)

2

\frac{1}{2} k y^{2}

3

\frac{1}{2} k (x^{2} + y^{2})

4

\frac{1}{2} k (x + y)^{2}

Explanation:

$W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x + y)^{2}$ $W = \frac{1}{2} k x^{2} - \frac{1}{2} k (x^{2} + y^{2} + 2 x y)$
$= \frac{1}{2} k y (- 2 x - y)$
The work done against elastic force is
$W_{e x t} = - W = \frac{k y}{2} (2 x + y)$

PHXI06:WORK ENERGY AND POWER

355614 A string of length $L$ and force constant $K$ is stretched to obtain extension $l$ . It is further stretched to obtain extension $l_{1}$ . The work done in second stretching is

1

\frac{1}{2} K l_{1}^{2}

2

\frac{1}{2} K (l_{1}^{2} - l^{2})

3

\frac{1}{2} K (l_{1}^{2} + l^{2})

4

\frac{1}{2} K l_{1} (2 l + l_{1})

Explanation:

Work done in stretching a string to obtain an extension $l$ is
$W_{1} = \frac{1}{2} K l$
Similarly, work done in stretching a string to obtain extension $l_{1}$ is
$W_{2} = \frac{1}{2} K l_{1}^{2}$
$∴$ Work done in second case
$= W_{2} - W_{1} = \frac{1}{2} K (l_{1}^{2} - l^{2})$

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