QBManager

Work, Energy and Power

148809 A small body of mass $500 g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2 m / s$ and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g = 10 m / s^{2}$ )

1

1 W

2

1.5 W

3

2 W

4

2.5 W

Explanation:

B Given, mass (m) $= 500 g = 0.5 kg$
Initial velocity of body $(u) = 2 m / s$
Coefficient of friction $(μ) = 0.3$
Frictional force $(f) = μ mg = 0.3 \times 0.5 \times 10 = 1.5 N$
$v_{avg} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 m / s$
Now,
$P_{avg} = f \times v_{avg}$
$P_{avg} = 1.5 \times 1$
$P_{avg} = 1.5 W$

TS- EAMCET-07.05.2018

Work, Energy and Power

148810 Consider a drop of rain water having mass $1 g$ falling from a height of $1 km$ . It hits the ground with a speed of $50 m / s$ . Take $g$ constant with a value of $10 m / s^{2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

1 (i)

- 10 J

, (ii)

- 8.25 J

2 (i)

1.25 J

, (ii)

- 8.25 J

3 (i)

100 J

, (ii)

8.75 J

4 (i)

10 J

, (ii)

- 8.75 J

Explanation:

D Given data,
Mass $(m) = 1 g$ or $1 \times 10^{- 3} kg$
Height $(h) = 1 km = 1000 m$
Velocity $(v) = 50 m / s$
$g = 10 m / s^{2}$
We know that,
Work done by gravitational force $(W_{g}) = mgh$
$W_{g} = 1 \times 10^{- 3} \times 10 \times 1000$
$W_{g} = 10 J$
Change in kinetic energy $(Δ$ K.E $) = \frac{1}{2} {mv}^{2} - 0$
$Δ K . E = \frac{1}{2} \times 1 \times 10^{- 3} \times 50 \times 50 - 0 = 1.25$
From Work-Energy theorem,
$W_{g} + W_{air resistance} = Δ K . E$
$10 + W_{air resistance} = 1.25$
$W_{air resistance} = 1.25 - 10$
$= - 8.75 J$

NEET- 2017

Work, Energy and Power

148811 A body of mass $m$ accelerates uniformly from rest to $v_{1}$ in time $t_{1}$ . The instantaneous power delivered to the body as a function of time

1

\frac{m v_{1}^{2} t}{t_{1}}

2

\frac{{mv}_{1} t}{t_{1}}

3

\frac{{mv}_{1} t^{2}}{t_{1}}

4

\frac{m v_{1}^{2} t}{t^{2}}

Explanation:

D Given,
mass $= m$ ,
Initial velocity $(v_{i}) = 0$ ,
Final velocity $(v_{f}) = v_{1}$
We know that,
Acceleration (a) $= \frac{Δ v}{Δ t} = \frac{v_{f} - v_{i}}{Δ t}$
$a = \frac{v_{1} - 0}{t_{1}} = \frac{v_{1}}{t_{1}}$
Apply equation of motion-
$v = u + at$
$v = 0 + at$
$v = \frac{v_{1}}{t_{1}} \times t$
Now,
$Power (P) = F . v$
$P = m . av$
$P = m \times (\frac{v_{1}}{t_{1}}) \times (\frac{v_{1}}{t_{1}}) t$
$P = \frac{{mv}_{1}^{2} t}{t_{1}^{2}}$

AMU-2017

Work, Energy and Power

148812 If a particle's position is given by $x = 4 - 12 t +$ $3 t^{2}$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t = 1 s$ ? Whether the particle is moving in $+ x$ direction or $- x$ direction?

1

- 6 m / s, + x

direction

2

- 6 m / s, - x

direction

3

6 m / s, + x

direction

4

4 m / s, - x

direction

Explanation:

B Given Relation, $x = 4 - 12 t + 3 t^{2}$
Velocity of the particle $(v) = \frac{d x}{d t}$
$v = \frac{d}{dt} (4 - 12 t + 3 t^{2})$
$v = - 12 + 6 t$
At $t = 1 s, v = - 12 + 6 \times 1$
$v = - 6 m / s, (- x direction)$

AMU-2017

Work, Energy and Power

148813 A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of $5 cm / s$ and the balloon shrinks completely in $2.5 s$ . The average force acting on the balloon will be

1 200 dyne

2 20 dyne

3 20 Newton

4 2000]**# dyne

Explanation:

B Given data,
Mass of balloon $(m) = 10$ gram
Velocity $(v) = 5 cm / s$
Time $(t) = 2.5 s$
Force $(F) =$ ?
We know that,
$F = m a$
$F = \frac{m v}{t}$
$F = \frac{10 \times 5}{2.5}$
$F = 20 dyne$

AMU-2017

Work, Energy and Power

148809 A small body of mass $500 g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2 m / s$ and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g = 10 m / s^{2}$ )

1

1 W

2

1.5 W

3

2 W

4

2.5 W

Explanation:

B Given, mass (m) $= 500 g = 0.5 kg$
Initial velocity of body $(u) = 2 m / s$
Coefficient of friction $(μ) = 0.3$
Frictional force $(f) = μ mg = 0.3 \times 0.5 \times 10 = 1.5 N$
$v_{avg} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 m / s$
Now,
$P_{avg} = f \times v_{avg}$
$P_{avg} = 1.5 \times 1$
$P_{avg} = 1.5 W$

TS- EAMCET-07.05.2018

Work, Energy and Power

148810 Consider a drop of rain water having mass $1 g$ falling from a height of $1 km$ . It hits the ground with a speed of $50 m / s$ . Take $g$ constant with a value of $10 m / s^{2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

1 (i)

- 10 J

, (ii)

- 8.25 J

2 (i)

1.25 J

, (ii)

- 8.25 J

3 (i)

100 J

, (ii)

8.75 J

4 (i)

10 J

, (ii)

- 8.75 J

Explanation:

D Given data,
Mass $(m) = 1 g$ or $1 \times 10^{- 3} kg$
Height $(h) = 1 km = 1000 m$
Velocity $(v) = 50 m / s$
$g = 10 m / s^{2}$
We know that,
Work done by gravitational force $(W_{g}) = mgh$
$W_{g} = 1 \times 10^{- 3} \times 10 \times 1000$
$W_{g} = 10 J$
Change in kinetic energy $(Δ$ K.E $) = \frac{1}{2} {mv}^{2} - 0$
$Δ K . E = \frac{1}{2} \times 1 \times 10^{- 3} \times 50 \times 50 - 0 = 1.25$
From Work-Energy theorem,
$W_{g} + W_{air resistance} = Δ K . E$
$10 + W_{air resistance} = 1.25$
$W_{air resistance} = 1.25 - 10$
$= - 8.75 J$

NEET- 2017

Work, Energy and Power

148811 A body of mass $m$ accelerates uniformly from rest to $v_{1}$ in time $t_{1}$ . The instantaneous power delivered to the body as a function of time

1

\frac{m v_{1}^{2} t}{t_{1}}

2

\frac{{mv}_{1} t}{t_{1}}

3

\frac{{mv}_{1} t^{2}}{t_{1}}

4

\frac{m v_{1}^{2} t}{t^{2}}

Explanation:

D Given,
mass $= m$ ,
Initial velocity $(v_{i}) = 0$ ,
Final velocity $(v_{f}) = v_{1}$
We know that,
Acceleration (a) $= \frac{Δ v}{Δ t} = \frac{v_{f} - v_{i}}{Δ t}$
$a = \frac{v_{1} - 0}{t_{1}} = \frac{v_{1}}{t_{1}}$
Apply equation of motion-
$v = u + at$
$v = 0 + at$
$v = \frac{v_{1}}{t_{1}} \times t$
Now,
$Power (P) = F . v$
$P = m . av$
$P = m \times (\frac{v_{1}}{t_{1}}) \times (\frac{v_{1}}{t_{1}}) t$
$P = \frac{{mv}_{1}^{2} t}{t_{1}^{2}}$

AMU-2017

Work, Energy and Power

148812 If a particle's position is given by $x = 4 - 12 t +$ $3 t^{2}$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t = 1 s$ ? Whether the particle is moving in $+ x$ direction or $- x$ direction?

1

- 6 m / s, + x

direction

2

- 6 m / s, - x

direction

3

6 m / s, + x

direction

4

4 m / s, - x

direction

Explanation:

B Given Relation, $x = 4 - 12 t + 3 t^{2}$
Velocity of the particle $(v) = \frac{d x}{d t}$
$v = \frac{d}{dt} (4 - 12 t + 3 t^{2})$
$v = - 12 + 6 t$
At $t = 1 s, v = - 12 + 6 \times 1$
$v = - 6 m / s, (- x direction)$

AMU-2017

Work, Energy and Power

148813 A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of $5 cm / s$ and the balloon shrinks completely in $2.5 s$ . The average force acting on the balloon will be

1 200 dyne

2 20 dyne

3 20 Newton

4 2000]**# dyne

Explanation:

B Given data,
Mass of balloon $(m) = 10$ gram
Velocity $(v) = 5 cm / s$
Time $(t) = 2.5 s$
Force $(F) =$ ?
We know that,
$F = m a$
$F = \frac{m v}{t}$
$F = \frac{10 \times 5}{2.5}$
$F = 20 dyne$

AMU-2017

Work, Energy and Power

148809 A small body of mass $500 g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2 m / s$ and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g = 10 m / s^{2}$ )

1

1 W

2

1.5 W

3

2 W

4

2.5 W

Explanation:

B Given, mass (m) $= 500 g = 0.5 kg$
Initial velocity of body $(u) = 2 m / s$
Coefficient of friction $(μ) = 0.3$
Frictional force $(f) = μ mg = 0.3 \times 0.5 \times 10 = 1.5 N$
$v_{avg} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 m / s$
Now,
$P_{avg} = f \times v_{avg}$
$P_{avg} = 1.5 \times 1$
$P_{avg} = 1.5 W$

TS- EAMCET-07.05.2018

Work, Energy and Power

148810 Consider a drop of rain water having mass $1 g$ falling from a height of $1 km$ . It hits the ground with a speed of $50 m / s$ . Take $g$ constant with a value of $10 m / s^{2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

1 (i)

- 10 J

, (ii)

- 8.25 J

2 (i)

1.25 J

, (ii)

- 8.25 J

3 (i)

100 J

, (ii)

8.75 J

4 (i)

10 J

, (ii)

- 8.75 J

Explanation:

D Given data,
Mass $(m) = 1 g$ or $1 \times 10^{- 3} kg$
Height $(h) = 1 km = 1000 m$
Velocity $(v) = 50 m / s$
$g = 10 m / s^{2}$
We know that,
Work done by gravitational force $(W_{g}) = mgh$
$W_{g} = 1 \times 10^{- 3} \times 10 \times 1000$
$W_{g} = 10 J$
Change in kinetic energy $(Δ$ K.E $) = \frac{1}{2} {mv}^{2} - 0$
$Δ K . E = \frac{1}{2} \times 1 \times 10^{- 3} \times 50 \times 50 - 0 = 1.25$
From Work-Energy theorem,
$W_{g} + W_{air resistance} = Δ K . E$
$10 + W_{air resistance} = 1.25$
$W_{air resistance} = 1.25 - 10$
$= - 8.75 J$

NEET- 2017

Work, Energy and Power

148811 A body of mass $m$ accelerates uniformly from rest to $v_{1}$ in time $t_{1}$ . The instantaneous power delivered to the body as a function of time

1

\frac{m v_{1}^{2} t}{t_{1}}

2

\frac{{mv}_{1} t}{t_{1}}

3

\frac{{mv}_{1} t^{2}}{t_{1}}

4

\frac{m v_{1}^{2} t}{t^{2}}

Explanation:

D Given,
mass $= m$ ,
Initial velocity $(v_{i}) = 0$ ,
Final velocity $(v_{f}) = v_{1}$
We know that,
Acceleration (a) $= \frac{Δ v}{Δ t} = \frac{v_{f} - v_{i}}{Δ t}$
$a = \frac{v_{1} - 0}{t_{1}} = \frac{v_{1}}{t_{1}}$
Apply equation of motion-
$v = u + at$
$v = 0 + at$
$v = \frac{v_{1}}{t_{1}} \times t$
Now,
$Power (P) = F . v$
$P = m . av$
$P = m \times (\frac{v_{1}}{t_{1}}) \times (\frac{v_{1}}{t_{1}}) t$
$P = \frac{{mv}_{1}^{2} t}{t_{1}^{2}}$

AMU-2017

Work, Energy and Power

148812 If a particle's position is given by $x = 4 - 12 t +$ $3 t^{2}$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t = 1 s$ ? Whether the particle is moving in $+ x$ direction or $- x$ direction?

1

- 6 m / s, + x

direction

2

- 6 m / s, - x

direction

3

6 m / s, + x

direction

4

4 m / s, - x

direction

Explanation:

B Given Relation, $x = 4 - 12 t + 3 t^{2}$
Velocity of the particle $(v) = \frac{d x}{d t}$
$v = \frac{d}{dt} (4 - 12 t + 3 t^{2})$
$v = - 12 + 6 t$
At $t = 1 s, v = - 12 + 6 \times 1$
$v = - 6 m / s, (- x direction)$

AMU-2017

Work, Energy and Power

148813 A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of $5 cm / s$ and the balloon shrinks completely in $2.5 s$ . The average force acting on the balloon will be

1 200 dyne

2 20 dyne

3 20 Newton

4 2000]**# dyne

Explanation:

B Given data,
Mass of balloon $(m) = 10$ gram
Velocity $(v) = 5 cm / s$
Time $(t) = 2.5 s$
Force $(F) =$ ?
We know that,
$F = m a$
$F = \frac{m v}{t}$
$F = \frac{10 \times 5}{2.5}$
$F = 20 dyne$

AMU-2017

Work, Energy and Power

148809 A small body of mass $500 g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2 m / s$ and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g = 10 m / s^{2}$ )

1

1 W

2

1.5 W

3

2 W

4

2.5 W

Explanation:

B Given, mass (m) $= 500 g = 0.5 kg$
Initial velocity of body $(u) = 2 m / s$
Coefficient of friction $(μ) = 0.3$
Frictional force $(f) = μ mg = 0.3 \times 0.5 \times 10 = 1.5 N$
$v_{avg} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 m / s$
Now,
$P_{avg} = f \times v_{avg}$
$P_{avg} = 1.5 \times 1$
$P_{avg} = 1.5 W$

TS- EAMCET-07.05.2018

Work, Energy and Power

148810 Consider a drop of rain water having mass $1 g$ falling from a height of $1 km$ . It hits the ground with a speed of $50 m / s$ . Take $g$ constant with a value of $10 m / s^{2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

1 (i)

- 10 J

, (ii)

- 8.25 J

2 (i)

1.25 J

, (ii)

- 8.25 J

3 (i)

100 J

, (ii)

8.75 J

4 (i)

10 J

, (ii)

- 8.75 J

Explanation:

D Given data,
Mass $(m) = 1 g$ or $1 \times 10^{- 3} kg$
Height $(h) = 1 km = 1000 m$
Velocity $(v) = 50 m / s$
$g = 10 m / s^{2}$
We know that,
Work done by gravitational force $(W_{g}) = mgh$
$W_{g} = 1 \times 10^{- 3} \times 10 \times 1000$
$W_{g} = 10 J$
Change in kinetic energy $(Δ$ K.E $) = \frac{1}{2} {mv}^{2} - 0$
$Δ K . E = \frac{1}{2} \times 1 \times 10^{- 3} \times 50 \times 50 - 0 = 1.25$
From Work-Energy theorem,
$W_{g} + W_{air resistance} = Δ K . E$
$10 + W_{air resistance} = 1.25$
$W_{air resistance} = 1.25 - 10$
$= - 8.75 J$

NEET- 2017

Work, Energy and Power

148811 A body of mass $m$ accelerates uniformly from rest to $v_{1}$ in time $t_{1}$ . The instantaneous power delivered to the body as a function of time

1

\frac{m v_{1}^{2} t}{t_{1}}

2

\frac{{mv}_{1} t}{t_{1}}

3

\frac{{mv}_{1} t^{2}}{t_{1}}

4

\frac{m v_{1}^{2} t}{t^{2}}

Explanation:

D Given,
mass $= m$ ,
Initial velocity $(v_{i}) = 0$ ,
Final velocity $(v_{f}) = v_{1}$
We know that,
Acceleration (a) $= \frac{Δ v}{Δ t} = \frac{v_{f} - v_{i}}{Δ t}$
$a = \frac{v_{1} - 0}{t_{1}} = \frac{v_{1}}{t_{1}}$
Apply equation of motion-
$v = u + at$
$v = 0 + at$
$v = \frac{v_{1}}{t_{1}} \times t$
Now,
$Power (P) = F . v$
$P = m . av$
$P = m \times (\frac{v_{1}}{t_{1}}) \times (\frac{v_{1}}{t_{1}}) t$
$P = \frac{{mv}_{1}^{2} t}{t_{1}^{2}}$

AMU-2017

Work, Energy and Power

148812 If a particle's position is given by $x = 4 - 12 t +$ $3 t^{2}$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t = 1 s$ ? Whether the particle is moving in $+ x$ direction or $- x$ direction?

1

- 6 m / s, + x

direction

2

- 6 m / s, - x

direction

3

6 m / s, + x

direction

4

4 m / s, - x

direction

Explanation:

B Given Relation, $x = 4 - 12 t + 3 t^{2}$
Velocity of the particle $(v) = \frac{d x}{d t}$
$v = \frac{d}{dt} (4 - 12 t + 3 t^{2})$
$v = - 12 + 6 t$
At $t = 1 s, v = - 12 + 6 \times 1$
$v = - 6 m / s, (- x direction)$

AMU-2017

Work, Energy and Power

148813 A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of $5 cm / s$ and the balloon shrinks completely in $2.5 s$ . The average force acting on the balloon will be

1 200 dyne

2 20 dyne

3 20 Newton

4 2000]**# dyne

Explanation:

B Given data,
Mass of balloon $(m) = 10$ gram
Velocity $(v) = 5 cm / s$
Time $(t) = 2.5 s$
Force $(F) =$ ?
We know that,
$F = m a$
$F = \frac{m v}{t}$
$F = \frac{10 \times 5}{2.5}$
$F = 20 dyne$

AMU-2017

Work, Energy and Power

148809 A small body of mass $500 g$ moves on a rough horizontal surface before finally stops. The initial velocity of the body is $2 m / s$ and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Take, $g = 10 m / s^{2}$ )

1

1 W

2

1.5 W

3

2 W

4

2.5 W

Explanation:

B Given, mass (m) $= 500 g = 0.5 kg$
Initial velocity of body $(u) = 2 m / s$
Coefficient of friction $(μ) = 0.3$
Frictional force $(f) = μ mg = 0.3 \times 0.5 \times 10 = 1.5 N$
$v_{avg} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 m / s$
Now,
$P_{avg} = f \times v_{avg}$
$P_{avg} = 1.5 \times 1$
$P_{avg} = 1.5 W$

TS- EAMCET-07.05.2018

Work, Energy and Power

148810 Consider a drop of rain water having mass $1 g$ falling from a height of $1 km$ . It hits the ground with a speed of $50 m / s$ . Take $g$ constant with a value of $10 m / s^{2}$ . The work done by the (i) gravitational force and the (ii) resistive force of air is

1 (i)

- 10 J

, (ii)

- 8.25 J

2 (i)

1.25 J

, (ii)

- 8.25 J

3 (i)

100 J

, (ii)

8.75 J

4 (i)

10 J

, (ii)

- 8.75 J

Explanation:

D Given data,
Mass $(m) = 1 g$ or $1 \times 10^{- 3} kg$
Height $(h) = 1 km = 1000 m$
Velocity $(v) = 50 m / s$
$g = 10 m / s^{2}$
We know that,
Work done by gravitational force $(W_{g}) = mgh$
$W_{g} = 1 \times 10^{- 3} \times 10 \times 1000$
$W_{g} = 10 J$
Change in kinetic energy $(Δ$ K.E $) = \frac{1}{2} {mv}^{2} - 0$
$Δ K . E = \frac{1}{2} \times 1 \times 10^{- 3} \times 50 \times 50 - 0 = 1.25$
From Work-Energy theorem,
$W_{g} + W_{air resistance} = Δ K . E$
$10 + W_{air resistance} = 1.25$
$W_{air resistance} = 1.25 - 10$
$= - 8.75 J$

NEET- 2017

Work, Energy and Power

148811 A body of mass $m$ accelerates uniformly from rest to $v_{1}$ in time $t_{1}$ . The instantaneous power delivered to the body as a function of time

1

\frac{m v_{1}^{2} t}{t_{1}}

2

\frac{{mv}_{1} t}{t_{1}}

3

\frac{{mv}_{1} t^{2}}{t_{1}}

4

\frac{m v_{1}^{2} t}{t^{2}}

Explanation:

D Given,
mass $= m$ ,
Initial velocity $(v_{i}) = 0$ ,
Final velocity $(v_{f}) = v_{1}$
We know that,
Acceleration (a) $= \frac{Δ v}{Δ t} = \frac{v_{f} - v_{i}}{Δ t}$
$a = \frac{v_{1} - 0}{t_{1}} = \frac{v_{1}}{t_{1}}$
Apply equation of motion-
$v = u + at$
$v = 0 + at$
$v = \frac{v_{1}}{t_{1}} \times t$
Now,
$Power (P) = F . v$
$P = m . av$
$P = m \times (\frac{v_{1}}{t_{1}}) \times (\frac{v_{1}}{t_{1}}) t$
$P = \frac{{mv}_{1}^{2} t}{t_{1}^{2}}$

AMU-2017

Work, Energy and Power

148812 If a particle's position is given by $x = 4 - 12 t +$ $3 t^{2}$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t = 1 s$ ? Whether the particle is moving in $+ x$ direction or $- x$ direction?

1

- 6 m / s, + x

direction

2

- 6 m / s, - x

direction

3

6 m / s, + x

direction

4

4 m / s, - x

direction

Explanation:

B Given Relation, $x = 4 - 12 t + 3 t^{2}$
Velocity of the particle $(v) = \frac{d x}{d t}$
$v = \frac{d}{dt} (4 - 12 t + 3 t^{2})$
$v = - 12 + 6 t$
At $t = 1 s, v = - 12 + 6 \times 1$
$v = - 6 m / s, (- x direction)$

AMU-2017

Work, Energy and Power

148813 A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of $5 cm / s$ and the balloon shrinks completely in $2.5 s$ . The average force acting on the balloon will be

1 200 dyne

2 20 dyne

3 20 Newton

4 2000]**# dyne

Explanation:

B Given data,
Mass of balloon $(m) = 10$ gram
Velocity $(v) = 5 cm / s$
Time $(t) = 2.5 s$
Force $(F) =$ ?
We know that,
$F = m a$
$F = \frac{m v}{t}$
$F = \frac{10 \times 5}{2.5}$
$F = 20 dyne$

AMU-2017