Question

Q1: An object of mass 3kg is moving on a rough surface with a velocity 16m/s its covers a distance of 20m before coming to rest. Calculate opposing force.​

Answer 1

Given:-

→Mass of the object = 3kg

→Initial velocity of the object = 16m/s

→Distance covered = 20m

To find:-

→Magnitude of opposing force

Solution:-

In this case:-

•Final velocity of the object will be zero as it is finally coming to rest.

Firstly, let's calculate the acceleration of the object.

By using the 3rd equation of motion, we get:-

=> v²-u² = 2as

=> 0-(16)² = 2×a×20

=> -256 = 40a

=> a = -256/40

=> a = -6.4m/s²

Hence, we have got the acceleration of the object as -6.4m/s².

Now, by using Newton's 2nd law of motion, we get:-

=> F = ma

=> F = 3(-6.4)

=> F = -19.2 N

[Here, -ve sign represents opposing force]

Thus, the opposing force is 19.2 N.