Question

A stone of 1kg is thrown with a velocity of 20 ms across a distance of a lake and comes to rest after traveling a distance of 50m what is the force of friction between the stone and the ice

Answer 1

Answer:

Given :

Initial velocity ,u = 20m/s^2

Final velocity,v = 0

Distance,s = 50 m

Mass of stone = 1 kg

To Find:

Force of friction between stone and ice

Solution:

$\bold\blue{\mathbb{{v}^{2} ={u}^{2}-2as}}$

$\bold\blue{\mathbb{》{0}^{2}={20}^{2}-2×a×50}}$

$\bold\blue{\mathbb{》0=400+100a}}$

$\bold\blue{\mathbb{》a=\frac{-400}{100}}}$

$\bold\blue{\mathbb{》a=-4m/{s}^{2}}}$

$\huge\bold\yellow{\mathbb{Now,}}$

Force = mass ×acceleration

or

F = m × a

$\bold\red{\mathbb{》F = 1- (-4) N}}$

$\boxed{\boxed{\boxed{\mathbb{》F = -4 N}}}}$

Thus, the force of friction between the stone and the ice is 4 newtons. The negative sign shows that this force opposes the motion of the stone.