Question

A car with a mass of `1000kg` initially moving at a rate of `20m/s` rolls to a stop. What is the work done by friction? What is the force of friction if it takes a distance of `20m` for the car to stop?

Answer 1

When a car with a mass of `1000kg` initially moving at a rate of `20ms^-1` rolls to a stop. As no brakes are applied all the energy is spent against friction only. Using Law of Conservation of energy.
Change in Kinetic energy of car
`Delta KE=KE_"Final"-KE_"Initial"`
`=>Delta KE=0-1/2xx1000xx20^2`
`=>Delta KE=-2.0xx10^5J`

This loss of energy is the work done by friction `=-2.0xx10^5J`

Using the kinematic expression and assuming that deceleration is constant
`v^2-u^2=2as`
we get acceleration as
`0^2-20^2=2axx20`
`=>a=-20^2/(2xx20)`
`=>a=-10ms^-2`

Using Newton's Second Law of motion, we get
Force `F=mxxa`
`F=1000xx(-10)`
`F=-10^4N`
`-ve` sign indicates that it is a retarding force and acts against the direction of motion.