posted by Lindsay on .
A 850.0-kg car travelling on a level road at 27.0 m/s (60.5 mi/hr) can stop, locking its wheels, in a distance of 61.0 m (200.1 ft).
(a) Find the size of the horizontal force which the car applies on the road while stopping.
(b) Find the stopping distance of that same car when it is traveling up a 18.1 deg slope, and it locks its wheels while traveling at 34.7 m/s (77.7 mi/hr). Assume that muk does not depend on the speed.
This is making no sense to me. Can I please have some help?
(a) (Force) x (stopping distance) = kinetic energy converted to heat
= (1/2) M V^2
Solve for F
(b) In this case, the initial kinetic energy is converted to both gravitational potential energy and heat.
The horizontal force will be whatever you get in Part (a), multiplied by cos 18.1 (0.9505). That is becasue the kinetic friction force is proportion to the normal component of the weight. Call this force F'
(Frictional work done) + (Potential energy increase) = (Initial kinetic energy)
F'*X + M g (X sin 18.1) = (1/2) M V^2
Solve for the new value of X.
The second term on the left is the increase in gravitational potential energy as the car sises up the slop.
I came up with 5079 N as my answer to a. Is that correct?
Yeppp n/m it is correct. Thanks a lot. :)