Question

A 6.00-kg block is sent up a ramp inclined at an angle =27.0∘ from the horizontal. It is given an initial velocity 0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is k=0.40 and the coefficient of static friction is s=0.70.

What distance along the ramp's surface does the block travel before it comes to a stop?

Answer 1

X = 8.2 m

Step-by-step explanation:

m = 4.25 kg, v0 = 15.0 m/s, v = 0 m/s

Since you have to push the block to overcome the static friction to start moving, the force required to set it in motion must be greater than the difference between the component of weight along the ramp and the static frictional force. This force with a minimum magnitude equal to the kinetic friction is required to keep the block in a constant motion. The normal force is N = mg cosθ

F ≥ μkN + μsN - mg sinθ = μk mg cosθ + μs mg cosθ - mg sinθ

=mg[( μk + μs)cosθ + sinθ]

=4.25 kgx9.8 m/s2[(0.368+0.663)cos31.5+sin31.5]

= 58.3754 N

Fnet = ma

ma = 58.3754, a = 13.735 m/s2

v2 = v02 - 2aX

02 = (15.0 m/s)2 -2(13.735m/s2)X

X = 8.2 m