Solve Final Speed of Ice Sliding Down Slope

[hyemal.zephyr]

Homework Statement

A block of ice with mass 2.0 kg slides 0.90 m down an inclined plane that slopes downward at an angle of 27° below the horizontal. If the block of ice starts from rest, what is its final speed? Friction can be neglected.

m = 2.0 kg
s = 0.90 m
θ = 27°

Homework Equations

W = Fs
W = ΔKE
ΔKE = 0.5m(v_f² - v_i²)

The Attempt at a Solution

Using the equations given, I began with the following:
Fs = 0.5m(v_f² - v_i²)

Then, because the ice is on a slope, I calculated the force acting on the block:
F = mgcosθ
F = (2.0 kg)(9.81 m/s²)cos(27°)
F = 17.48 N

Plugging all of the variables into the equation, now:
(17.48 N)(0.90 m) = 0.5(2.0 kg)(v_f² - (0 m/s)²)
v_f = 3.97 m/s

Apparently, though, this is incorrect. I'm not sure why.

As a side note, I know how to solve the problem using kinematics, but this topic is introducing work, so we have to solve the problem as such.

 

[Doc Al]

Then, because the ice is on a slope, I calculated the force acting on the block:
F = mgcosθ

That's the component of the weight perpendicular to the surface; you want the component parallel to the surface.

 

[hyemal.zephyr]

Yes, I got 2.83 m/s, which is apparently the correct answer. Thank you.

Though, why is that, exactly? It being the perpendicular component, that is. I've understood that as the parallel component and I've always seemed to do well with it.

 

[Doc Al]

Though, why is that, exactly? It being the perpendicular component, that is. I've understood that as the parallel component and I've always seemed to do well with it.

You need to review how to find components. See: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

 

[hyemal.zephyr]

Oh, I see what I've been doing incorrectly. Hah, I feel like a moron now.

Thank you very much.