A 8.75-kg block is sent up a ramp inclined at an angle θ = 30.5° 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 = 0.353 and the coefficient of static friction is μs = 0.639. How far up the ramp (in the direction along the ramp) does the block go before it comes to a stop?

Weight on the block: [tex]W = m \cdot g = 8.75 \times 9.81 = 85.8375\;\text{N}[/tex].
Consider the weight on the block as two components: along the slope and normal to the slope. Magnitude of the component along the slope: [tex]W_\text{along the slope} = W \cdot \sin{\theta} = 43.5658\;\text{N}[/tex].
Normal force on the block, which is the same as the component of weight normal to the slope: [tex]N = W_\text{normal to the slope} = W \cdot \cos{\theta} = 73.9601\;\text{N}[/tex].
Kinetic friction: [tex]F_\text{kinetic friction} = \mu_k\cdot N = 26.1079\;\text{N}[/tex].
Maximum Static friction: [tex]F_\text{static friction, max} = \mu_s\cdot N = 47.2605\;\text{N}[/tex].

[tex]F_\text{static friction, max} > W_\text{along the slope}[/tex]. As a result, the block won't slip downwards after it comes to a stop.

As the block moves upward:

Net force on the block [tex]\sum F = F_\text{kinetic friction} +W_\text{along the slope} = -69.6737\;\text{N}[/tex].
Acceleration of the block: [tex]{a} = \dfrac{{F}}{m} = -7.96271\;\text{m}\cdot\text{s}^{-2}[/tex].

Initial velocity [tex]v_0 = 15.0\;\text{m}\cdot\text{s}^{-1}[/tex].
Final velocity of the block [tex]v_1 = 0[/tex] as it comes to a stop, given that it doesn't slip downwards.

How far does the block move in this process?

[tex]x = \dfrac{{v_1}^{2}-{v_0}^{2}}{2a} = 14.1\;\text{m}[/tex] (3 sig. fig.).