Question

13, A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming the body to boot initially at rest, find (a) its velocity when the force stops acting (b) the distance covered in 10 after the force starts acting.​

Answer 1

Given :-

• Force = 4 N
• Mass = 2 kg
• Time = 4 s

To find :-

• Velocity when the force stops acting.
• Distance covered in 10s after the force starts acting.

Solution :-

a)

We have to find velocity when the force stops acting, it is given that the force stops acting after 4 seconds, So we have to find velocity after 4 sec.

For calculating velocity, we first need to find acceleration -

➩ Force = Mass × Acceleration

➩ 4 = 2 × a

➩ a = 4/2

➩ a = 2 m/s²

Calculating Final velocity -

• Initial velocity, u = 0 m/s
• Time, t = 4 sec
• Acceleration, a = 2 m/s²

Substituting the value in 1st equation of motion :-

➩ v = u + at

➩ v = 0 + 4 × 2

➩ v = 8 m/s

Final velocity = 8 m/s

ㅤ

ㅤ

b)

For calculating distance for 10s we have to solve for 2 different case. The force is acting on 4 sec and then the force is not acting, so we have to first solve for 4 sec and then next 6 sec.

Case 1 -

• u = 0 m/s
• a = 2 m/s²
• t = 4 sec

Substituting the value in 2nd equation of motion :-

➩ $\rm s_1 = ut + \frac{1}{2} at^2$

➩ $\rm s_1 = 0 + \frac{1}{ \cancel 2} \times \cancel 2 \times 4^2$

➩ $\rm s_1 = 16 m$

ㅤ

Case 2 -

Force acting is 0, So the acceleration also equals to 0.

The final velocity for 4 sec, will now be the inital velocity for 6 sec.

• a = 0 m/s²
• t = 6 sec
• u = 8 m/s

➩ $\rm s_2 = ut + \frac{1}{2} at^2$

➩ $\rm s_2 = 6 \times 8 + 0$

➩ $\rm s_2 = 48m$

ㅤ

ㅤ

ㅤ

Total distance travelled = 48 + 16 m

= 64 m

ㅤ

Distance covered = 64 m