Question

A force of 10N acts on a body of mass 2kg. If the body was initially at rest, calculate the velocity gained by it in 5 seconds. ​

Answer 1

Answer:

$\huge\mathfrak\pink{Answer}$

Explanation:

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the body

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a=

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= m

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF =

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3s

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/s

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/sStep 3: Change in momentum

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/sStep 3: Change in momentumChange in momentum = Final momentum − Initial momentum

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/sStep 3: Change in momentumChange in momentum = Final momentum − Initial momentum =mv−mu=m(v−u)

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/sStep 3: Change in momentumChange in momentum = Final momentum − Initial momentum =mv−mu=m(v−u) =2kg (15 m/s−0)

Given: F=10 N,m=2 kg,t=3s, Initial velocity, u=0Step 1: Acceleration of the bodyF=ma⇒a= mF = 2 Kg10 N =5m/s 2 Step 2: Velocity acquired by the body after 3sAs acceleration is constant, therefore applying equation of motion:v=u+at=0+5×3=15m/sStep 3: Change in momentumChange in momentum = Final momentum − Initial momentum =mv−mu=m(v−u) =2kg (15 m/s−0) =30 kg-m/s

Answer 2

Answer:

A force of 10N acts on a body of mass 2kg. If the body was initially at rest, The velocity gained by it in 5 seconds will be $25\ m/s$

Explanation:

The first equation motion is given by,

$v = u+at$                   ...(1)

Where,

u -  Initial velocity

v -  Final velocity

t  -  Time

a -  Acceleration

Acceleration is the ratio of change in velocity and time.

That is,   $a = \frac{v-u}{t}$        ...(2)

Newton's second law of motion is given by the expression,

$F = ma$

Where

F - Force

m - Mass of the body

Substitute equation(2) in (1)           

$F = m(\frac{v-u}{t})$                  ...(3)

From the question,

$F = 10\ N$

$m = 2\ kg$

$u = 0\ m/s$

$t = 5\ s$

Substitute these values into equation (3)

$10 = 2(\frac{v-0}{5})$

$\frac{10}{2}= \frac{v}{5}$

$v = 25\ m/s$