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

 

[Nickie]

I would say that your solution appears to be correct. You have correctly used the equations for work, kinetic friction, and change in kinetic energy to determine the block's final velocity after sliding on the rough surface. Then, you have used the equation for potential energy to calculate the height that the block will reach on the ramp before coming to rest. Your calculations and reasoning seem sound and logical. However, without knowing the specific context or purpose of this problem, it is difficult to say for certain if this is the "right" way to answer it. There may be other factors or variables that need to be considered, or different approaches that could also yield a correct solution. As a scientist, it is important to always question and critically evaluate our methods and results, and to be open to different perspectives and solutions.