How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

Question

A 4.25-kg block is sent up a ramp inclined at an angle θ = 31.5° from the horizontal. It is given an initial velocity v0 = 15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is μk = 0.368 and the coefficient of static friction is μs = 0.663. How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

Answer 1

Hi Payton,

 

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                              *       *       *

                          *               **

                              *       **μ_k = 0.368

                                  **  μ_s = 0.663

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                        * θ = 31.5°

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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 ≥ μ_kN + μ_sN - mg sinθ = μ_k mg cosθ + μ_s mg cosθ - mg sinθ

                                    =mg[( μ_k + μ_s)cosθ + sinθ]

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

                                    = 58.3754 N

F_net = 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