A stone of mass 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a take and it comes to rest after travelling a distance of 50 m . What is the magnitude of the force opposing the motion of the stone ?

Explanation

` u= 20 m//s`
`v = 0`
` s= 50 m`
Mass , m= 1 kg
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) + 2 xx a xx 50`
` 100 = -400 `
` a = -(400)/100 = - 4 m//s^(2)`
Acceleration ` a = -4 m//s^(2) " "` (Negatie sign shows that speed of the stone decreases, i.e retardation )
Now, F = ma = `1 "kg " xx (-4) m//a^(2)`
` = -4 "Kg"-m//s^(2) `
` = -4 N`
Thus , force of friction between the stone and the ice is -4 N.
The negative value of force shows that the frictional force acts in a direction opposite to the direction of motion.