How High Does the Block Go on the Ramp?

[kushlar]

Homework Statement

A block of mass 5.00 kg is given an initial velocity of 8.0 m/s. It travels 5.0 m over a rough surface with μ_{k} = 0.250. It then slides up a ramp that rises at a 30° angle above the horizontal. How high does it go before coming to rest

Homework Equations

W = FΔs
F_{k} = μ_{k} N
Δk = 1/2 m v_{f}^{2} - 1/2 m v_{i}^{2}
v = √2gh

The Attempt at a Solution

W = μ_{k} Δs = (0.25)(5)(9.8)(5) = 61.25 J

-61.25 = Δk
-61.25 = 1/2 m v_{f}^{2} - 1/2 m v_{i}^{2}
-61.25 = 2.5 v_{f}^{2} - (0.5)(5)(8^{2}
v_{f} = √((-61.25+160)/2.5)

v = √2gh
h = v^{2}/2gh
h = 6.28^{2}/(2)(9.8)
h = 2.01m


Is this the right way to answer this question?

 

[SHISHKABOB]

is the ramp frictionless or is it also a rough surface? It looks like you assumed that it is frictionless, but I guess it makes sense to do that since it doesn't specify that in the problem.

But yes, assuming the ramp is frictionless, it looks like you did it right.

 

[kushlar]

I believe the ramp is frictionless but the one thing that bothers me about my solution is the lack of the 30° angle in my calculations.

 

[SHISHKABOB]

yeah that would be a piece of extraneous information for this problem

since you are using conservation of energy, the path that the object takes doesn't matter, only the initial and final states