Question

A car of mass 1800kg moving with a speed of 10m/s is brought to rest after a covering a distance of 50m. calculate the force acting on the car.

Answer 1

u = 10 m/s
v = 0
s = 50 m
a = ?

Applying 3rd kinematic equation---
v^2 - u^2 = 2as
(0)^2 - (10)^2 = 2 x 50 x a
-100 = 100a
a = -1 m/s^2

Now m = 1800 kg
a = -1m/s^2

F = ma
F = 1800 x 1
F = -1800 N

The force is acting in the opposite direction to the motion of the car.

Hope This Helps You

Answer 2

Using third law of 1d kinematics,
v^2-u^2=2as
Given, s=50m; v=0; u=10m/s
Then , v^2-u^2=2as
0-(10)^2=2a(50)
a= -1m/s
Force=mass×acceleration, (or) Force=mass×deceleration
Therefore, deceleration= -1m/s^2
Deceleration= -(acceleration)
Force=mass×deceleration
=1800×1
=1800N
Force will be acting in opposite direction to displacement of the car

Hope it helps
Pls mark it as the brainliest