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