QBManager

Work, Energy and Power

149130 A cable in the form of a spiral roll (shown in the figure) has a linear density $ρ$ . It is uncoiled at a uniform speed $v$ , if the total length of the cable is $L$ . The work done in uncoiling the cable is

1

ρ L v^{2} / 4

2

ρ L v^{2} / 2

3

ρ {Lv}^{2} / 3

4

ρ L v^{2}

Explanation:

B Given that,
Initial speed of the coil $(u) = 0 m / s$
Linear density $= ρ$
Final speed $= v$ (uncoiling speed $)$
Length of the cable $= L$
Mass of coil $= ρ L$
So,
Work done, $W = Δ KE = K_{f} - K_{i}$
$W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$
$W = \frac{1}{2} m v^{2} - 0$
$W = \frac{1}{2} \times (ρ L) (v)^{2}$
$W = \frac{1}{2} ρ {Lv}^{2}$

JCECE-2015

Work, Energy and Power

149131 A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is:

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

A Given that,
Minimum height of swing from the earth's surface,
$H_{1} = 0.75 m$
Maximum height of swing from earth's surface,
$H_{2} = 2 m$
From energy conservation rule,

Kinetic energy $=$ Potential energy
$\frac{1}{2} {mv}_{max}^{2} = mg (H_{2} - H_{1})$
$\frac{1}{2} {mv}_{max}^{2} = mg (2 - 0.75)$
$v_{max} = \sqrt{2 \times 10 \times 1.25}$
$v_{max} = \sqrt{25}$
$v_{max} = 5 m / s$

JCECE-2004

Work, Energy and Power

149132 Two bodies of masses $0.1 kg$ and $0.4 kg$ move towards each other with the velocities $1 m / s$ and $0.1 m / s$ respectively. After collision they stick together. In 10 sec the combined mass travels:

1

120 m

2

0.12 m

3

12 m

4

1.2 m

Explanation:

D Given that,
Mass of first body $(m_{1}) = 0.1 kg$
Mass of second body $(m_{2}) = 0.4 kg$
Velocity before collision $(v_{1}) = 1 m / s$
Velocity before collision moving opposite direction $(v_{2})$ $= - 0.1 m / s$
$t = 10 s$
According to conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
$(0.1) (1) + (0.4) (- 0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$v = 0.12 m / s$
$∴$ Distance, $s = vt$
$s = 0.12 \times 10 = 1.2 m$

AIIMS-2010

Work, Energy and Power

149133 If the escape speed of a projectile on Earth's surface is $11.2 {kms}^{- 1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

1

56.63 {kms}^{- 1}

2

33 {kms}^{- 1}

3

39 {kms}^{- 1}

4

31.7 {kms}^{- 1}

Explanation:

D Given,
Escape velocity $(v_{e}) = 11.2 {kms}^{- 1}$
Projected speed $(v_{p}) = 3 v_{e}$
Mass of projectile $= m$
Suppose energy of projectile far away from the earth $=$ $v_{1}$
$Total energy = \frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2}$
Total Energy of the projectile far away from the earth
$= \frac{1}{2} {mv}_{1}^{2} (At infinite, P.E. = 0)$
Applying law of conservation of energy,
$\frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2} = \frac{1}{2} {mv}_{1}^{2}$
$v_{1} = \sqrt{v_{p}^{2} - v_{e}^{2}}$
$v_{1} = \sqrt{{(3 v_{e})}^{2} - {(v_{e})}^{2}}$
$v_{1} = \sqrt{8} v_{e}$
$v_{1} = \sqrt{8} \times 11.2$
$v_{1} = 31.6736 km / s$
$v_{1} ≃ 31.7 {kms}^{- 1}$
Speed of the body at for away from the earth is $= 31.7 {kms}^{- 1}$

BCECE-2015

Work, Energy and Power

149134 A spring of constant $5 \times 10^{3} N / m$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is-

1

6.25 Nm

2

12.5 Nm

3

18.75 Nm

4

25.00 Nm

Explanation:

C Given that,
Spring constant $(k) = 5 \times 10^{3} N / m$
$x_{1} = 5 cm = 0.05 m, x_{2} = x_{1} + 5 cm$
$x_{2} = 10 cm = 0.1 m$
We know that, $W = Δ PE$
$W = {PE}_{2} - {PE}_{1}$
Now, ${PE}_{1} = \frac{1}{2} {kx}_{1}^{2}$
${PE}_{1} = \frac{1}{2} \times 5 \times 10^{3} (0.05)^{2} = 6.25 J$
${PE}_{2} = \frac{1}{2} {kx}_{2}^{2}$
${PE}_{2} = \frac{1}{2} \times 5 \times 10^{3} (0.1)^{2} = 25 J$
$W = {PE}_{2} - {PE}_{1} = 25 - 6.25 = 18.75 J$

BCECE-2011

Work, Energy and Power

149130 A cable in the form of a spiral roll (shown in the figure) has a linear density $ρ$ . It is uncoiled at a uniform speed $v$ , if the total length of the cable is $L$ . The work done in uncoiling the cable is

1

ρ L v^{2} / 4

2

ρ L v^{2} / 2

3

ρ {Lv}^{2} / 3

4

ρ L v^{2}

Explanation:

B Given that,
Initial speed of the coil $(u) = 0 m / s$
Linear density $= ρ$
Final speed $= v$ (uncoiling speed $)$
Length of the cable $= L$
Mass of coil $= ρ L$
So,
Work done, $W = Δ KE = K_{f} - K_{i}$
$W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$
$W = \frac{1}{2} m v^{2} - 0$
$W = \frac{1}{2} \times (ρ L) (v)^{2}$
$W = \frac{1}{2} ρ {Lv}^{2}$

JCECE-2015

Work, Energy and Power

149131 A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is:

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

A Given that,
Minimum height of swing from the earth's surface,
$H_{1} = 0.75 m$
Maximum height of swing from earth's surface,
$H_{2} = 2 m$
From energy conservation rule,

Kinetic energy $=$ Potential energy
$\frac{1}{2} {mv}_{max}^{2} = mg (H_{2} - H_{1})$
$\frac{1}{2} {mv}_{max}^{2} = mg (2 - 0.75)$
$v_{max} = \sqrt{2 \times 10 \times 1.25}$
$v_{max} = \sqrt{25}$
$v_{max} = 5 m / s$

JCECE-2004

Work, Energy and Power

149132 Two bodies of masses $0.1 kg$ and $0.4 kg$ move towards each other with the velocities $1 m / s$ and $0.1 m / s$ respectively. After collision they stick together. In 10 sec the combined mass travels:

1

120 m

2

0.12 m

3

12 m

4

1.2 m

Explanation:

D Given that,
Mass of first body $(m_{1}) = 0.1 kg$
Mass of second body $(m_{2}) = 0.4 kg$
Velocity before collision $(v_{1}) = 1 m / s$
Velocity before collision moving opposite direction $(v_{2})$ $= - 0.1 m / s$
$t = 10 s$
According to conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
$(0.1) (1) + (0.4) (- 0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$v = 0.12 m / s$
$∴$ Distance, $s = vt$
$s = 0.12 \times 10 = 1.2 m$

AIIMS-2010

Work, Energy and Power

149133 If the escape speed of a projectile on Earth's surface is $11.2 {kms}^{- 1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

1

56.63 {kms}^{- 1}

2

33 {kms}^{- 1}

3

39 {kms}^{- 1}

4

31.7 {kms}^{- 1}

Explanation:

D Given,
Escape velocity $(v_{e}) = 11.2 {kms}^{- 1}$
Projected speed $(v_{p}) = 3 v_{e}$
Mass of projectile $= m$
Suppose energy of projectile far away from the earth $=$ $v_{1}$
$Total energy = \frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2}$
Total Energy of the projectile far away from the earth
$= \frac{1}{2} {mv}_{1}^{2} (At infinite, P.E. = 0)$
Applying law of conservation of energy,
$\frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2} = \frac{1}{2} {mv}_{1}^{2}$
$v_{1} = \sqrt{v_{p}^{2} - v_{e}^{2}}$
$v_{1} = \sqrt{{(3 v_{e})}^{2} - {(v_{e})}^{2}}$
$v_{1} = \sqrt{8} v_{e}$
$v_{1} = \sqrt{8} \times 11.2$
$v_{1} = 31.6736 km / s$
$v_{1} ≃ 31.7 {kms}^{- 1}$
Speed of the body at for away from the earth is $= 31.7 {kms}^{- 1}$

BCECE-2015

Work, Energy and Power

149134 A spring of constant $5 \times 10^{3} N / m$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is-

1

6.25 Nm

2

12.5 Nm

3

18.75 Nm

4

25.00 Nm

Explanation:

C Given that,
Spring constant $(k) = 5 \times 10^{3} N / m$
$x_{1} = 5 cm = 0.05 m, x_{2} = x_{1} + 5 cm$
$x_{2} = 10 cm = 0.1 m$
We know that, $W = Δ PE$
$W = {PE}_{2} - {PE}_{1}$
Now, ${PE}_{1} = \frac{1}{2} {kx}_{1}^{2}$
${PE}_{1} = \frac{1}{2} \times 5 \times 10^{3} (0.05)^{2} = 6.25 J$
${PE}_{2} = \frac{1}{2} {kx}_{2}^{2}$
${PE}_{2} = \frac{1}{2} \times 5 \times 10^{3} (0.1)^{2} = 25 J$
$W = {PE}_{2} - {PE}_{1} = 25 - 6.25 = 18.75 J$

BCECE-2011

Work, Energy and Power

149130 A cable in the form of a spiral roll (shown in the figure) has a linear density $ρ$ . It is uncoiled at a uniform speed $v$ , if the total length of the cable is $L$ . The work done in uncoiling the cable is

1

ρ L v^{2} / 4

2

ρ L v^{2} / 2

3

ρ {Lv}^{2} / 3

4

ρ L v^{2}

Explanation:

B Given that,
Initial speed of the coil $(u) = 0 m / s$
Linear density $= ρ$
Final speed $= v$ (uncoiling speed $)$
Length of the cable $= L$
Mass of coil $= ρ L$
So,
Work done, $W = Δ KE = K_{f} - K_{i}$
$W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$
$W = \frac{1}{2} m v^{2} - 0$
$W = \frac{1}{2} \times (ρ L) (v)^{2}$
$W = \frac{1}{2} ρ {Lv}^{2}$

JCECE-2015

Work, Energy and Power

149131 A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is:

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

A Given that,
Minimum height of swing from the earth's surface,
$H_{1} = 0.75 m$
Maximum height of swing from earth's surface,
$H_{2} = 2 m$
From energy conservation rule,

Kinetic energy $=$ Potential energy
$\frac{1}{2} {mv}_{max}^{2} = mg (H_{2} - H_{1})$
$\frac{1}{2} {mv}_{max}^{2} = mg (2 - 0.75)$
$v_{max} = \sqrt{2 \times 10 \times 1.25}$
$v_{max} = \sqrt{25}$
$v_{max} = 5 m / s$

JCECE-2004

Work, Energy and Power

149132 Two bodies of masses $0.1 kg$ and $0.4 kg$ move towards each other with the velocities $1 m / s$ and $0.1 m / s$ respectively. After collision they stick together. In 10 sec the combined mass travels:

1

120 m

2

0.12 m

3

12 m

4

1.2 m

Explanation:

D Given that,
Mass of first body $(m_{1}) = 0.1 kg$
Mass of second body $(m_{2}) = 0.4 kg$
Velocity before collision $(v_{1}) = 1 m / s$
Velocity before collision moving opposite direction $(v_{2})$ $= - 0.1 m / s$
$t = 10 s$
According to conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
$(0.1) (1) + (0.4) (- 0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$v = 0.12 m / s$
$∴$ Distance, $s = vt$
$s = 0.12 \times 10 = 1.2 m$

AIIMS-2010

Work, Energy and Power

149133 If the escape speed of a projectile on Earth's surface is $11.2 {kms}^{- 1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

1

56.63 {kms}^{- 1}

2

33 {kms}^{- 1}

3

39 {kms}^{- 1}

4

31.7 {kms}^{- 1}

Explanation:

D Given,
Escape velocity $(v_{e}) = 11.2 {kms}^{- 1}$
Projected speed $(v_{p}) = 3 v_{e}$
Mass of projectile $= m$
Suppose energy of projectile far away from the earth $=$ $v_{1}$
$Total energy = \frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2}$
Total Energy of the projectile far away from the earth
$= \frac{1}{2} {mv}_{1}^{2} (At infinite, P.E. = 0)$
Applying law of conservation of energy,
$\frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2} = \frac{1}{2} {mv}_{1}^{2}$
$v_{1} = \sqrt{v_{p}^{2} - v_{e}^{2}}$
$v_{1} = \sqrt{{(3 v_{e})}^{2} - {(v_{e})}^{2}}$
$v_{1} = \sqrt{8} v_{e}$
$v_{1} = \sqrt{8} \times 11.2$
$v_{1} = 31.6736 km / s$
$v_{1} ≃ 31.7 {kms}^{- 1}$
Speed of the body at for away from the earth is $= 31.7 {kms}^{- 1}$

BCECE-2015

Work, Energy and Power

149134 A spring of constant $5 \times 10^{3} N / m$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is-

1

6.25 Nm

2

12.5 Nm

3

18.75 Nm

4

25.00 Nm

Explanation:

C Given that,
Spring constant $(k) = 5 \times 10^{3} N / m$
$x_{1} = 5 cm = 0.05 m, x_{2} = x_{1} + 5 cm$
$x_{2} = 10 cm = 0.1 m$
We know that, $W = Δ PE$
$W = {PE}_{2} - {PE}_{1}$
Now, ${PE}_{1} = \frac{1}{2} {kx}_{1}^{2}$
${PE}_{1} = \frac{1}{2} \times 5 \times 10^{3} (0.05)^{2} = 6.25 J$
${PE}_{2} = \frac{1}{2} {kx}_{2}^{2}$
${PE}_{2} = \frac{1}{2} \times 5 \times 10^{3} (0.1)^{2} = 25 J$
$W = {PE}_{2} - {PE}_{1} = 25 - 6.25 = 18.75 J$

BCECE-2011

Work, Energy and Power

149130 A cable in the form of a spiral roll (shown in the figure) has a linear density $ρ$ . It is uncoiled at a uniform speed $v$ , if the total length of the cable is $L$ . The work done in uncoiling the cable is

1

ρ L v^{2} / 4

2

ρ L v^{2} / 2

3

ρ {Lv}^{2} / 3

4

ρ L v^{2}

Explanation:

B Given that,
Initial speed of the coil $(u) = 0 m / s$
Linear density $= ρ$
Final speed $= v$ (uncoiling speed $)$
Length of the cable $= L$
Mass of coil $= ρ L$
So,
Work done, $W = Δ KE = K_{f} - K_{i}$
$W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$
$W = \frac{1}{2} m v^{2} - 0$
$W = \frac{1}{2} \times (ρ L) (v)^{2}$
$W = \frac{1}{2} ρ {Lv}^{2}$

JCECE-2015

Work, Energy and Power

149131 A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is:

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

A Given that,
Minimum height of swing from the earth's surface,
$H_{1} = 0.75 m$
Maximum height of swing from earth's surface,
$H_{2} = 2 m$
From energy conservation rule,

Kinetic energy $=$ Potential energy
$\frac{1}{2} {mv}_{max}^{2} = mg (H_{2} - H_{1})$
$\frac{1}{2} {mv}_{max}^{2} = mg (2 - 0.75)$
$v_{max} = \sqrt{2 \times 10 \times 1.25}$
$v_{max} = \sqrt{25}$
$v_{max} = 5 m / s$

JCECE-2004

Work, Energy and Power

149132 Two bodies of masses $0.1 kg$ and $0.4 kg$ move towards each other with the velocities $1 m / s$ and $0.1 m / s$ respectively. After collision they stick together. In 10 sec the combined mass travels:

1

120 m

2

0.12 m

3

12 m

4

1.2 m

Explanation:

D Given that,
Mass of first body $(m_{1}) = 0.1 kg$
Mass of second body $(m_{2}) = 0.4 kg$
Velocity before collision $(v_{1}) = 1 m / s$
Velocity before collision moving opposite direction $(v_{2})$ $= - 0.1 m / s$
$t = 10 s$
According to conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
$(0.1) (1) + (0.4) (- 0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$v = 0.12 m / s$
$∴$ Distance, $s = vt$
$s = 0.12 \times 10 = 1.2 m$

AIIMS-2010

Work, Energy and Power

149133 If the escape speed of a projectile on Earth's surface is $11.2 {kms}^{- 1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

1

56.63 {kms}^{- 1}

2

33 {kms}^{- 1}

3

39 {kms}^{- 1}

4

31.7 {kms}^{- 1}

Explanation:

D Given,
Escape velocity $(v_{e}) = 11.2 {kms}^{- 1}$
Projected speed $(v_{p}) = 3 v_{e}$
Mass of projectile $= m$
Suppose energy of projectile far away from the earth $=$ $v_{1}$
$Total energy = \frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2}$
Total Energy of the projectile far away from the earth
$= \frac{1}{2} {mv}_{1}^{2} (At infinite, P.E. = 0)$
Applying law of conservation of energy,
$\frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2} = \frac{1}{2} {mv}_{1}^{2}$
$v_{1} = \sqrt{v_{p}^{2} - v_{e}^{2}}$
$v_{1} = \sqrt{{(3 v_{e})}^{2} - {(v_{e})}^{2}}$
$v_{1} = \sqrt{8} v_{e}$
$v_{1} = \sqrt{8} \times 11.2$
$v_{1} = 31.6736 km / s$
$v_{1} ≃ 31.7 {kms}^{- 1}$
Speed of the body at for away from the earth is $= 31.7 {kms}^{- 1}$

BCECE-2015

Work, Energy and Power

149134 A spring of constant $5 \times 10^{3} N / m$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is-

1

6.25 Nm

2

12.5 Nm

3

18.75 Nm

4

25.00 Nm

Explanation:

C Given that,
Spring constant $(k) = 5 \times 10^{3} N / m$
$x_{1} = 5 cm = 0.05 m, x_{2} = x_{1} + 5 cm$
$x_{2} = 10 cm = 0.1 m$
We know that, $W = Δ PE$
$W = {PE}_{2} - {PE}_{1}$
Now, ${PE}_{1} = \frac{1}{2} {kx}_{1}^{2}$
${PE}_{1} = \frac{1}{2} \times 5 \times 10^{3} (0.05)^{2} = 6.25 J$
${PE}_{2} = \frac{1}{2} {kx}_{2}^{2}$
${PE}_{2} = \frac{1}{2} \times 5 \times 10^{3} (0.1)^{2} = 25 J$
$W = {PE}_{2} - {PE}_{1} = 25 - 6.25 = 18.75 J$

BCECE-2011

Work, Energy and Power

149130 A cable in the form of a spiral roll (shown in the figure) has a linear density $ρ$ . It is uncoiled at a uniform speed $v$ , if the total length of the cable is $L$ . The work done in uncoiling the cable is

1

ρ L v^{2} / 4

2

ρ L v^{2} / 2

3

ρ {Lv}^{2} / 3

4

ρ L v^{2}

Explanation:

B Given that,
Initial speed of the coil $(u) = 0 m / s$
Linear density $= ρ$
Final speed $= v$ (uncoiling speed $)$
Length of the cable $= L$
Mass of coil $= ρ L$
So,
Work done, $W = Δ KE = K_{f} - K_{i}$
$W = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$
$W = \frac{1}{2} m v^{2} - 0$
$W = \frac{1}{2} \times (ρ L) (v)^{2}$
$W = \frac{1}{2} ρ {Lv}^{2}$

JCECE-2015

Work, Energy and Power

149131 A child is swinging a swing. Minimum and maximum heights of swing from earth's surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is:

1

5 m / s

2

10 m / s

3

15 m / s

4

20 m / s

Explanation:

A Given that,
Minimum height of swing from the earth's surface,
$H_{1} = 0.75 m$
Maximum height of swing from earth's surface,
$H_{2} = 2 m$
From energy conservation rule,

Kinetic energy $=$ Potential energy
$\frac{1}{2} {mv}_{max}^{2} = mg (H_{2} - H_{1})$
$\frac{1}{2} {mv}_{max}^{2} = mg (2 - 0.75)$
$v_{max} = \sqrt{2 \times 10 \times 1.25}$
$v_{max} = \sqrt{25}$
$v_{max} = 5 m / s$

JCECE-2004

Work, Energy and Power

149132 Two bodies of masses $0.1 kg$ and $0.4 kg$ move towards each other with the velocities $1 m / s$ and $0.1 m / s$ respectively. After collision they stick together. In 10 sec the combined mass travels:

1

120 m

2

0.12 m

3

12 m

4

1.2 m

Explanation:

D Given that,
Mass of first body $(m_{1}) = 0.1 kg$
Mass of second body $(m_{2}) = 0.4 kg$
Velocity before collision $(v_{1}) = 1 m / s$
Velocity before collision moving opposite direction $(v_{2})$ $= - 0.1 m / s$
$t = 10 s$
According to conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v$
$(0.1) (1) + (0.4) (- 0.1) = (0.1 + 0.4) v$
$0.1 - 0.04 = 0.5 v$
$v = 0.12 m / s$
$∴$ Distance, $s = vt$
$s = 0.12 \times 10 = 1.2 m$

AIIMS-2010

Work, Energy and Power

149133 If the escape speed of a projectile on Earth's surface is $11.2 {kms}^{- 1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth.

1

56.63 {kms}^{- 1}

2

33 {kms}^{- 1}

3

39 {kms}^{- 1}

4

31.7 {kms}^{- 1}

Explanation:

D Given,
Escape velocity $(v_{e}) = 11.2 {kms}^{- 1}$
Projected speed $(v_{p}) = 3 v_{e}$
Mass of projectile $= m$
Suppose energy of projectile far away from the earth $=$ $v_{1}$
$Total energy = \frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2}$
Total Energy of the projectile far away from the earth
$= \frac{1}{2} {mv}_{1}^{2} (At infinite, P.E. = 0)$
Applying law of conservation of energy,
$\frac{1}{2} {mv}_{p}^{2} - \frac{1}{2} {mv}_{e}^{2} = \frac{1}{2} {mv}_{1}^{2}$
$v_{1} = \sqrt{v_{p}^{2} - v_{e}^{2}}$
$v_{1} = \sqrt{{(3 v_{e})}^{2} - {(v_{e})}^{2}}$
$v_{1} = \sqrt{8} v_{e}$
$v_{1} = \sqrt{8} \times 11.2$
$v_{1} = 31.6736 km / s$
$v_{1} ≃ 31.7 {kms}^{- 1}$
Speed of the body at for away from the earth is $= 31.7 {kms}^{- 1}$

BCECE-2015

Work, Energy and Power

149134 A spring of constant $5 \times 10^{3} N / m$ is stretched initially by $5 cm$ from the unstretched position. Then the work required to stretch it further by another $5 cm$ is-

1

6.25 Nm

2

12.5 Nm

3

18.75 Nm

4

25.00 Nm

Explanation:

C Given that,
Spring constant $(k) = 5 \times 10^{3} N / m$
$x_{1} = 5 cm = 0.05 m, x_{2} = x_{1} + 5 cm$
$x_{2} = 10 cm = 0.1 m$
We know that, $W = Δ PE$
$W = {PE}_{2} - {PE}_{1}$
Now, ${PE}_{1} = \frac{1}{2} {kx}_{1}^{2}$
${PE}_{1} = \frac{1}{2} \times 5 \times 10^{3} (0.05)^{2} = 6.25 J$
${PE}_{2} = \frac{1}{2} {kx}_{2}^{2}$
${PE}_{2} = \frac{1}{2} \times 5 \times 10^{3} (0.1)^{2} = 25 J$
$W = {PE}_{2} - {PE}_{1} = 25 - 6.25 = 18.75 J$

BCECE-2011