Question

Question #aeffb

Answer 1

Let the mass of the car be `m` and its initial velcity of rolling be `v`

So its initial KE `E_k=1/2mv^2`

Given that the coefficient of rolling fricion is `mu_r=0.02`

Rolling frictionoal force on the car `F_r=mu_rmg`. If the car coasts `s` m before stopping, then Initial KE will be equal in magnitude of the work done against friction.

So `1/2mv^2=mu_rmgxxs`

`=>s=v^2/(2mu_rg)`

Now `v=70"km/hr"=(70xx10^3)/3600=700/36"m/s"`

`g=9.8"m/"s^2`

Therefore

`s=(700/36)^2/(2xx0.02xx9.8)~~964.5m`