Question

8. A body of mass 20 kg is moving with a speed of 5 ms-1. Calculate the distance travelled by the
body before coming to rest when a constant retarding force of 40 N is applied on it. 2​

Answer 1

Answer:

6.25m

Explanation:

Given :

Mass of the body= 20 kg

Initial velocity = 5 m/s

Final velocity = 0 m/s

Force = -40N [retardation force]

To Find :

Distance travelled = ?

Solution :

According to the third equation of motion.

$\sf{}v^2-u^2=2as$

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

We don’t know acceleration so let’s find out acceleration first

We know F = ma

Here,

F is the force applied

m is the mass

a is the acceleration

$\sf{}\implies -40 = 20 \times a$

$\sf{}\implies a = -\dfrac{40}{20}$

$\sf{}\therefore a = -2m/s^2$

Thus,distance,

$\sf{}v^2-u^2=2as$

$\sf{}\implies 0-25=2\times -2\times s$

$\sf{}\implies -25=-4 s$

$\sf{}\implies\dfrac{-25}{-4}=s$

$\sf{}\therefore s=6.25m$

Therefore,distance travelled is 6.25m