# 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?

Mar 11, 2017

When a car with a mass of $1000 k g$ initially moving at a rate of $20 m {s}^{-} 1$ 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
$\Delta K E = K {E}_{\text{Final"-KE_"Initial}}$
$\implies \Delta K E = 0 - \frac{1}{2} \times 1000 \times {20}^{2}$
$\implies \Delta K E = - 2.0 \times {10}^{5} J$

This loss of energy is the work done by friction $= - 2.0 \times {10}^{5} J$

Using the kinematic expression and assuming that deceleration is constant
${v}^{2} - {u}^{2} = 2 a s$
we get acceleration as
${0}^{2} - {20}^{2} = 2 a \times 20$
$\implies a = - {20}^{2} / \left(2 \times 20\right)$
$\implies a = - 10 m {s}^{-} 2$

Using Newton's Second Law of motion, we get
Force $F = m \times a$
$F = 1000 \times \left(- 10\right)$
$F = - {10}^{4} N$
$- v e$ sign indicates that it is a retarding force and acts against the direction of motion.