A stone of mass 1 kg is thrown with a velocity of `20ms^(-1)` across the frozen surface of a lake and it comes to rest after travelling a distance of 50m. What is the magnitude of the force opposing the motion of the stone?

Explanation

`u=20m//s`
`v=0`
`s=50m`
Mass, `m=1kg`
To calculate force, we have the formula F = ma, but we have to first calculate acceleration a.
Using the third equation of motion, i.e.,
`v^(2)=u^(2)+2as`
`(0)^(2)=(20)^(2)+2xxaxx50`
`100a= -400`
`a= -(400)/(100)= -4 m//s^(2)`
Acceleration `a=-4m//s^(2)` (-ve sign shows that speed of the stone decreases, i.e., retardation)
Now, `F=ma=1kg xx (-4)m//s^(2)`
`=-4 "kg-m"//s^(2)`
`=-4N`
Thus, force of friction between the stone and the ice is -4N.
The negative value of force shows that the frictional force acts in a direction opposite to the direction of motion.