Question

A car can decelerate at -3.40 m/s2 without skidding when coming to rest on a level road. What would be the magnitude of its deceleration if the road were inclined at 14° uphill? Assume the same static friction coefficient.

Answer 1

well, the braking force is F=ma=-3.40m
and that is equal to mu*mg
so mu=3.40/9,8
Now going on a hill, the normal force is less (normal force=mg*Cos14), so braking force is mu*normal= 3.40/9.8*mgCos14=3.40*mCos14

That is the breaking force.

However, on a hill, gravity also adds a force down the hill mgSin14
This force aids braking (going up hill), or resists braking.

Going downhill:
3.40*m*Cos14-mgSin14=ma

a= 3.40Cos14-9.8Sin14

Going uphill:
3.40*m*Cos14+9.8sin14=ma
a= 3.40Cos14+9.8Sin14

Answer 2

4.7m/s