physics help?

Question

a 2100 kg car starts from rest at the top of a 5.0 m long driveway that is inclined at 20 degrees with the horizontal. If an average friction force of 4000 N impedes the motion, find the speed of the car at the bottom of the driveway.

Answer 1

Ok, what I understands, is that the car is on the top of a plane making, 20 degrees with the horizontal.

You can use the work - energy theorem, where :

Final energy minus initial energy = minus (work done by the force of friction).

initial force = potential energy = m*g*5

let's consider g = 10m/s^2 >>> gravity

2100*10*5 = 105000 Joules

final energy : m*v^2/2, where "v" is the speed of the car at the bottom.

work donde bye the force = 4000*distance

calculating the distance :

It's a plane making 20 degrees, so, the distance "d" :

d*sin20 = 5 >>> d = 14.6 m

then : the work done by the friction = 14.6*4000 = 58400 Joules

Using the theorem :

mv^2/2 - 105000 = -58400

mv^2 */ 2 = 46600

2100v^2 / 2 = 46600

v = 6.6 m/s

hope that might help you

Answer 2

sq.root of (21000cos20-4000)/2100 m/s