QBManager

Work, Energy and Power

148777 A force of $4 N$ acts on a $10 kg$ body initially at rest. Let $W_{1}$ is work done by force during 0 $\leq t \leq 1$ s. Likewise $W_{2}$ is the work done by force during $1 s \leq t \leq 2 s$ , where $t$ is time in second. The ratio $\frac{W_{2}}{W_{1}}$ is

1 1

2 2

3

\frac{1}{2}

4 3

Explanation:

D Given, $m = 10 kg, F = 4 N$
We know that,
Force, $F = m . a$
$\frac{F}{m} = a$
$\frac{4}{10} = a \Rightarrow a = 0.4 {ms}^{- 2}$
For $0 \leq t \leq 1 s$ , displacement $s_{1}$ is,
$s_{1} = {ut}_{1} + \frac{1}{2} {at}_{1}^{2}$
$s_{1} = 0 \times 1 + \frac{1}{2} \times 0.4 \times 1 = 0.2 m$
Work done $(W_{1}) = F \times s_{1} = 4 \times 0.2 = 0.8 J$ ....(i)
Now, for $1 s \leq t \leq 2 s$
$t_{2} = 1 s$
Velocity of the body after $t_{1}$ , we get-
$v = u + {at}_{1}$
$v = 0 + 0.4 \times 1 = 0.4 {ms}^{- 1}$
Displacement $s_{2}$ is,
$s_{2} = {vt}_{2} + \frac{1}{2} {at}_{2}^{2}$
$s_{2} = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times 1^{2} = 0.6 m$
$W_{2} = F . s_{2}$
$W_{2} = 4 \times 0.6 = 2.4 J$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{2.4}{0.8} = 3$

TS- EAMCET-11.09.2020

Work, Energy and Power

148778 A boy pushes a toy box $2.0 m$ along the floor by means of a force of $10 N$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

1

6 J

2

8 J

3

10 J

4

12 J

Explanation:

C Given,
$F = 10 N$
$d = 2 m$
$θ = 60^{\circ}$
We know that,
$W = \vec{F} \cdot \vec{d}$
$W = Fd \cos θ$
$W = Fd \cos 60^{\circ}$
$W = 10 \times 2 \times \frac{1}{2}, {∵ \cos 60^{\circ} = \frac{1}{2}}$
$W = 10 J$

BITSAT-2020

Work, Energy and Power

148779 A man pushes the wall and fails to displace it. He does

1 negative work

2 positive but not maximum work

3 no work at all

4 maximum work

Explanation:

C When a man pushes the wall but he failed to displace it. So, he does no work at all.
So, the displacement $(s) = 0$
$W = F . s$
$W = F \times 0$
$W = 0$
Hence, no work done at all.

AP EAMCET (21.09.2020) Shift-II

Work, Energy and Power

148780 An automatic gun fires 360 bullets per minute with a speed of $360 km / h$ . If each bullet weight $20 g$ , the power of the gun is

1

75 W

2

150 W

3

300 W

4

600 W

Explanation:

D Given,
Speed $(v) = 360 km / h$ ,
Mass $(m) = 20 g = 20 \times 10^{- 3} kg$
$v = 360 \times \frac{5}{18} = 100 m / s$
No. of bullets $(n) = 360 / \min$
Power $= \frac{Total energy}{time}$
$\begin{array}{r} = \frac{n \times (K . E) of each bullet}{time} = \frac{n \times \frac{1}{2} {mv}^{2}}{t} \\ = \frac{360 \times \frac{1}{2} \times (20 \times 10^{- 3}) (100)^{2}}{60} \\ = \frac{360 \times 10^{5} \times 10^{- 3}}{60} \\ = 6 \times 100 \\ P = 600 W \end{array}$

AP EAMCET (21.09.2020) Shift-I

Work, Energy and Power

148781 A force $F = 20 + 10 y$ acts on a particle in $y$ direction, where $F$ is in Newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

1

5 J

2

25 J

3

20 J

4

30 J

Explanation:

B Given that,
$F = 20 + 10 y$
Work done by variable force is
$W = \int_{y_{1}}^{y_{2}} F \cdot dy$
Here $y_{1} = 0, y_{2} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$W = 20 + 5$
$W = 25 J$

NEET -2019

Work, Energy and Power

148777 A force of $4 N$ acts on a $10 kg$ body initially at rest. Let $W_{1}$ is work done by force during 0 $\leq t \leq 1$ s. Likewise $W_{2}$ is the work done by force during $1 s \leq t \leq 2 s$ , where $t$ is time in second. The ratio $\frac{W_{2}}{W_{1}}$ is

1 1

2 2

3

\frac{1}{2}

4 3

Explanation:

D Given, $m = 10 kg, F = 4 N$
We know that,
Force, $F = m . a$
$\frac{F}{m} = a$
$\frac{4}{10} = a \Rightarrow a = 0.4 {ms}^{- 2}$
For $0 \leq t \leq 1 s$ , displacement $s_{1}$ is,
$s_{1} = {ut}_{1} + \frac{1}{2} {at}_{1}^{2}$
$s_{1} = 0 \times 1 + \frac{1}{2} \times 0.4 \times 1 = 0.2 m$
Work done $(W_{1}) = F \times s_{1} = 4 \times 0.2 = 0.8 J$ ....(i)
Now, for $1 s \leq t \leq 2 s$
$t_{2} = 1 s$
Velocity of the body after $t_{1}$ , we get-
$v = u + {at}_{1}$
$v = 0 + 0.4 \times 1 = 0.4 {ms}^{- 1}$
Displacement $s_{2}$ is,
$s_{2} = {vt}_{2} + \frac{1}{2} {at}_{2}^{2}$
$s_{2} = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times 1^{2} = 0.6 m$
$W_{2} = F . s_{2}$
$W_{2} = 4 \times 0.6 = 2.4 J$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{2.4}{0.8} = 3$

TS- EAMCET-11.09.2020

Work, Energy and Power

148778 A boy pushes a toy box $2.0 m$ along the floor by means of a force of $10 N$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

1

6 J

2

8 J

3

10 J

4

12 J

Explanation:

C Given,
$F = 10 N$
$d = 2 m$
$θ = 60^{\circ}$
We know that,
$W = \vec{F} \cdot \vec{d}$
$W = Fd \cos θ$
$W = Fd \cos 60^{\circ}$
$W = 10 \times 2 \times \frac{1}{2}, {∵ \cos 60^{\circ} = \frac{1}{2}}$
$W = 10 J$

BITSAT-2020

Work, Energy and Power

148779 A man pushes the wall and fails to displace it. He does

1 negative work

2 positive but not maximum work

3 no work at all

4 maximum work

Explanation:

C When a man pushes the wall but he failed to displace it. So, he does no work at all.
So, the displacement $(s) = 0$
$W = F . s$
$W = F \times 0$
$W = 0$
Hence, no work done at all.

AP EAMCET (21.09.2020) Shift-II

Work, Energy and Power

148780 An automatic gun fires 360 bullets per minute with a speed of $360 km / h$ . If each bullet weight $20 g$ , the power of the gun is

1

75 W

2

150 W

3

300 W

4

600 W

Explanation:

D Given,
Speed $(v) = 360 km / h$ ,
Mass $(m) = 20 g = 20 \times 10^{- 3} kg$
$v = 360 \times \frac{5}{18} = 100 m / s$
No. of bullets $(n) = 360 / \min$
Power $= \frac{Total energy}{time}$
$\begin{array}{r} = \frac{n \times (K . E) of each bullet}{time} = \frac{n \times \frac{1}{2} {mv}^{2}}{t} \\ = \frac{360 \times \frac{1}{2} \times (20 \times 10^{- 3}) (100)^{2}}{60} \\ = \frac{360 \times 10^{5} \times 10^{- 3}}{60} \\ = 6 \times 100 \\ P = 600 W \end{array}$

AP EAMCET (21.09.2020) Shift-I

Work, Energy and Power

148781 A force $F = 20 + 10 y$ acts on a particle in $y$ direction, where $F$ is in Newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

1

5 J

2

25 J

3

20 J

4

30 J

Explanation:

B Given that,
$F = 20 + 10 y$
Work done by variable force is
$W = \int_{y_{1}}^{y_{2}} F \cdot dy$
Here $y_{1} = 0, y_{2} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$W = 20 + 5$
$W = 25 J$

NEET -2019

Work, Energy and Power

148777 A force of $4 N$ acts on a $10 kg$ body initially at rest. Let $W_{1}$ is work done by force during 0 $\leq t \leq 1$ s. Likewise $W_{2}$ is the work done by force during $1 s \leq t \leq 2 s$ , where $t$ is time in second. The ratio $\frac{W_{2}}{W_{1}}$ is

1 1

2 2

3

\frac{1}{2}

4 3

Explanation:

D Given, $m = 10 kg, F = 4 N$
We know that,
Force, $F = m . a$
$\frac{F}{m} = a$
$\frac{4}{10} = a \Rightarrow a = 0.4 {ms}^{- 2}$
For $0 \leq t \leq 1 s$ , displacement $s_{1}$ is,
$s_{1} = {ut}_{1} + \frac{1}{2} {at}_{1}^{2}$
$s_{1} = 0 \times 1 + \frac{1}{2} \times 0.4 \times 1 = 0.2 m$
Work done $(W_{1}) = F \times s_{1} = 4 \times 0.2 = 0.8 J$ ....(i)
Now, for $1 s \leq t \leq 2 s$
$t_{2} = 1 s$
Velocity of the body after $t_{1}$ , we get-
$v = u + {at}_{1}$
$v = 0 + 0.4 \times 1 = 0.4 {ms}^{- 1}$
Displacement $s_{2}$ is,
$s_{2} = {vt}_{2} + \frac{1}{2} {at}_{2}^{2}$
$s_{2} = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times 1^{2} = 0.6 m$
$W_{2} = F . s_{2}$
$W_{2} = 4 \times 0.6 = 2.4 J$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{2.4}{0.8} = 3$

TS- EAMCET-11.09.2020

Work, Energy and Power

148778 A boy pushes a toy box $2.0 m$ along the floor by means of a force of $10 N$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

1

6 J

2

8 J

3

10 J

4

12 J

Explanation:

C Given,
$F = 10 N$
$d = 2 m$
$θ = 60^{\circ}$
We know that,
$W = \vec{F} \cdot \vec{d}$
$W = Fd \cos θ$
$W = Fd \cos 60^{\circ}$
$W = 10 \times 2 \times \frac{1}{2}, {∵ \cos 60^{\circ} = \frac{1}{2}}$
$W = 10 J$

BITSAT-2020

Work, Energy and Power

148779 A man pushes the wall and fails to displace it. He does

1 negative work

2 positive but not maximum work

3 no work at all

4 maximum work

Explanation:

C When a man pushes the wall but he failed to displace it. So, he does no work at all.
So, the displacement $(s) = 0$
$W = F . s$
$W = F \times 0$
$W = 0$
Hence, no work done at all.

AP EAMCET (21.09.2020) Shift-II

Work, Energy and Power

148780 An automatic gun fires 360 bullets per minute with a speed of $360 km / h$ . If each bullet weight $20 g$ , the power of the gun is

1

75 W

2

150 W

3

300 W

4

600 W

Explanation:

D Given,
Speed $(v) = 360 km / h$ ,
Mass $(m) = 20 g = 20 \times 10^{- 3} kg$
$v = 360 \times \frac{5}{18} = 100 m / s$
No. of bullets $(n) = 360 / \min$
Power $= \frac{Total energy}{time}$
$\begin{array}{r} = \frac{n \times (K . E) of each bullet}{time} = \frac{n \times \frac{1}{2} {mv}^{2}}{t} \\ = \frac{360 \times \frac{1}{2} \times (20 \times 10^{- 3}) (100)^{2}}{60} \\ = \frac{360 \times 10^{5} \times 10^{- 3}}{60} \\ = 6 \times 100 \\ P = 600 W \end{array}$

AP EAMCET (21.09.2020) Shift-I

Work, Energy and Power

148781 A force $F = 20 + 10 y$ acts on a particle in $y$ direction, where $F$ is in Newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

1

5 J

2

25 J

3

20 J

4

30 J

Explanation:

B Given that,
$F = 20 + 10 y$
Work done by variable force is
$W = \int_{y_{1}}^{y_{2}} F \cdot dy$
Here $y_{1} = 0, y_{2} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$W = 20 + 5$
$W = 25 J$

NEET -2019

Work, Energy and Power

148777 A force of $4 N$ acts on a $10 kg$ body initially at rest. Let $W_{1}$ is work done by force during 0 $\leq t \leq 1$ s. Likewise $W_{2}$ is the work done by force during $1 s \leq t \leq 2 s$ , where $t$ is time in second. The ratio $\frac{W_{2}}{W_{1}}$ is

1 1

2 2

3

\frac{1}{2}

4 3

Explanation:

D Given, $m = 10 kg, F = 4 N$
We know that,
Force, $F = m . a$
$\frac{F}{m} = a$
$\frac{4}{10} = a \Rightarrow a = 0.4 {ms}^{- 2}$
For $0 \leq t \leq 1 s$ , displacement $s_{1}$ is,
$s_{1} = {ut}_{1} + \frac{1}{2} {at}_{1}^{2}$
$s_{1} = 0 \times 1 + \frac{1}{2} \times 0.4 \times 1 = 0.2 m$
Work done $(W_{1}) = F \times s_{1} = 4 \times 0.2 = 0.8 J$ ....(i)
Now, for $1 s \leq t \leq 2 s$
$t_{2} = 1 s$
Velocity of the body after $t_{1}$ , we get-
$v = u + {at}_{1}$
$v = 0 + 0.4 \times 1 = 0.4 {ms}^{- 1}$
Displacement $s_{2}$ is,
$s_{2} = {vt}_{2} + \frac{1}{2} {at}_{2}^{2}$
$s_{2} = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times 1^{2} = 0.6 m$
$W_{2} = F . s_{2}$
$W_{2} = 4 \times 0.6 = 2.4 J$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{2.4}{0.8} = 3$

TS- EAMCET-11.09.2020

Work, Energy and Power

148778 A boy pushes a toy box $2.0 m$ along the floor by means of a force of $10 N$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

1

6 J

2

8 J

3

10 J

4

12 J

Explanation:

C Given,
$F = 10 N$
$d = 2 m$
$θ = 60^{\circ}$
We know that,
$W = \vec{F} \cdot \vec{d}$
$W = Fd \cos θ$
$W = Fd \cos 60^{\circ}$
$W = 10 \times 2 \times \frac{1}{2}, {∵ \cos 60^{\circ} = \frac{1}{2}}$
$W = 10 J$

BITSAT-2020

Work, Energy and Power

148779 A man pushes the wall and fails to displace it. He does

1 negative work

2 positive but not maximum work

3 no work at all

4 maximum work

Explanation:

C When a man pushes the wall but he failed to displace it. So, he does no work at all.
So, the displacement $(s) = 0$
$W = F . s$
$W = F \times 0$
$W = 0$
Hence, no work done at all.

AP EAMCET (21.09.2020) Shift-II

Work, Energy and Power

148780 An automatic gun fires 360 bullets per minute with a speed of $360 km / h$ . If each bullet weight $20 g$ , the power of the gun is

1

75 W

2

150 W

3

300 W

4

600 W

Explanation:

D Given,
Speed $(v) = 360 km / h$ ,
Mass $(m) = 20 g = 20 \times 10^{- 3} kg$
$v = 360 \times \frac{5}{18} = 100 m / s$
No. of bullets $(n) = 360 / \min$
Power $= \frac{Total energy}{time}$
$\begin{array}{r} = \frac{n \times (K . E) of each bullet}{time} = \frac{n \times \frac{1}{2} {mv}^{2}}{t} \\ = \frac{360 \times \frac{1}{2} \times (20 \times 10^{- 3}) (100)^{2}}{60} \\ = \frac{360 \times 10^{5} \times 10^{- 3}}{60} \\ = 6 \times 100 \\ P = 600 W \end{array}$

AP EAMCET (21.09.2020) Shift-I

Work, Energy and Power

148781 A force $F = 20 + 10 y$ acts on a particle in $y$ direction, where $F$ is in Newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

1

5 J

2

25 J

3

20 J

4

30 J

Explanation:

B Given that,
$F = 20 + 10 y$
Work done by variable force is
$W = \int_{y_{1}}^{y_{2}} F \cdot dy$
Here $y_{1} = 0, y_{2} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$W = 20 + 5$
$W = 25 J$

NEET -2019

Work, Energy and Power

148777 A force of $4 N$ acts on a $10 kg$ body initially at rest. Let $W_{1}$ is work done by force during 0 $\leq t \leq 1$ s. Likewise $W_{2}$ is the work done by force during $1 s \leq t \leq 2 s$ , where $t$ is time in second. The ratio $\frac{W_{2}}{W_{1}}$ is

1 1

2 2

3

\frac{1}{2}

4 3

Explanation:

D Given, $m = 10 kg, F = 4 N$
We know that,
Force, $F = m . a$
$\frac{F}{m} = a$
$\frac{4}{10} = a \Rightarrow a = 0.4 {ms}^{- 2}$
For $0 \leq t \leq 1 s$ , displacement $s_{1}$ is,
$s_{1} = {ut}_{1} + \frac{1}{2} {at}_{1}^{2}$
$s_{1} = 0 \times 1 + \frac{1}{2} \times 0.4 \times 1 = 0.2 m$
Work done $(W_{1}) = F \times s_{1} = 4 \times 0.2 = 0.8 J$ ....(i)
Now, for $1 s \leq t \leq 2 s$
$t_{2} = 1 s$
Velocity of the body after $t_{1}$ , we get-
$v = u + {at}_{1}$
$v = 0 + 0.4 \times 1 = 0.4 {ms}^{- 1}$
Displacement $s_{2}$ is,
$s_{2} = {vt}_{2} + \frac{1}{2} {at}_{2}^{2}$
$s_{2} = 0.4 \times 1 + \frac{1}{2} \times 0.4 \times 1^{2} = 0.6 m$
$W_{2} = F . s_{2}$
$W_{2} = 4 \times 0.6 = 2.4 J$
Dividing equation (ii) by (i), we get-
$\frac{W_{2}}{W_{1}} = \frac{2.4}{0.8} = 3$

TS- EAMCET-11.09.2020

Work, Energy and Power

148778 A boy pushes a toy box $2.0 m$ along the floor by means of a force of $10 N$ directed downward at an angle of $60^{\circ}$ to the horizontal. The work done by the boy is

1

6 J

2

8 J

3

10 J

4

12 J

Explanation:

C Given,
$F = 10 N$
$d = 2 m$
$θ = 60^{\circ}$
We know that,
$W = \vec{F} \cdot \vec{d}$
$W = Fd \cos θ$
$W = Fd \cos 60^{\circ}$
$W = 10 \times 2 \times \frac{1}{2}, {∵ \cos 60^{\circ} = \frac{1}{2}}$
$W = 10 J$

BITSAT-2020

Work, Energy and Power

148779 A man pushes the wall and fails to displace it. He does

1 negative work

2 positive but not maximum work

3 no work at all

4 maximum work

Explanation:

C When a man pushes the wall but he failed to displace it. So, he does no work at all.
So, the displacement $(s) = 0$
$W = F . s$
$W = F \times 0$
$W = 0$
Hence, no work done at all.

AP EAMCET (21.09.2020) Shift-II

Work, Energy and Power

148780 An automatic gun fires 360 bullets per minute with a speed of $360 km / h$ . If each bullet weight $20 g$ , the power of the gun is

1

75 W

2

150 W

3

300 W

4

600 W

Explanation:

D Given,
Speed $(v) = 360 km / h$ ,
Mass $(m) = 20 g = 20 \times 10^{- 3} kg$
$v = 360 \times \frac{5}{18} = 100 m / s$
No. of bullets $(n) = 360 / \min$
Power $= \frac{Total energy}{time}$
$\begin{array}{r} = \frac{n \times (K . E) of each bullet}{time} = \frac{n \times \frac{1}{2} {mv}^{2}}{t} \\ = \frac{360 \times \frac{1}{2} \times (20 \times 10^{- 3}) (100)^{2}}{60} \\ = \frac{360 \times 10^{5} \times 10^{- 3}}{60} \\ = 6 \times 100 \\ P = 600 W \end{array}$

AP EAMCET (21.09.2020) Shift-I

Work, Energy and Power

148781 A force $F = 20 + 10 y$ acts on a particle in $y$ direction, where $F$ is in Newton and $y$ in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

1

5 J

2

25 J

3

20 J

4

30 J

Explanation:

B Given that,
$F = 20 + 10 y$
Work done by variable force is
$W = \int_{y_{1}}^{y_{2}} F \cdot dy$
Here $y_{1} = 0, y_{2} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$W = 20 + 5$
$W = 25 J$

NEET -2019

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