Find magnitude of breaking force to stop car of mass 1.2T in 20m when its travelling 60k/hr. down an incline of angle arcin(1/40)
A couple of ways of doing this:
a) Using the standard Newton's equations (I personally don't like this method, but it's the "official" way of doing it), this is the way you'll have to answer the question in an exam:
m = 1.2T = 1200 Kg
v = 60 km/h = 50/3 m/s
To overcome gravity, a force of mg*sin@ = 294 N is required (up the slope).
In order to slow the car down in 20m, an additional acceleration of (v^2 - u^2)/ 2s = 125/18 m/s^2 is required
-> additional force required =
= 125/18 * 1200 N
25000/3 N (up the slope).
Adding these two forces gives 294 + 25000/3 N = 25882/3 N = 8627.33 N (up the slope)
b) Using the work-energy theorem (not in the syllabus, but simpler & more elegant in my opinion):
The mechanical energy (E) of the car is the sum of the kinetic energy and gravitational potential energy (E = (1/2)*mv^2 + mglsin@), so the initial mechanical energy is E = 500000/3 + 5880 J = 517640 J (taking the bottom of the 20m ramp to be the level of zero potential). The final mechanical energy is zero, so the force required to stop the car is (change in mechanical energy)/distance = 517640/20 N = 25882/3 N = 8627.33 N (up the slope)
You probably need to understand the first method, but not the second.
