Kinetic Energy problem

[PiRsq]

I don't know why I am getting a different answer but here is the question:


A toboggan is initially moving at a constant velocity along a snowy horizontal surface. Ignore friction. When a pulling force is applied parallel to the ground over a certain distance, the kinetic energy increases by 47%. By what percentage would the kinetic energy have changed if the pulling force had been at an angle of 38° above the horizontal?


A few things:

Ekf = Final kinetic energy
Eki = Initial kinetic energy
m=mass
vf=final velocity
vi=initial velocity
Now:

Case 1 (where the force applied is parallel to the ground)

2Facos0(mvf-mvi)=2(100 x 0.47)

mvf-mvi=(100 x 0.47)/Fa


Case 2 (where the force is at 38° to the horizontal)

2Facos38(mvf-mvi)=2(100 x y)

mvf-mvi= (100 x y )/FaCos38


Now equating the two equations:

100 x 0.47 = (100 x y)/cos38

Solving for y I got y=0.37

So the increase is ~ 10%, and the answer is 16%. What did I do wrong?

 

[Doc Al]

I'm not sure what you're calculating. Think of it this way:

(horizontal component of F)x (horizontal displacement) = KE(f)-KE(i)

The initial KE(i) is, of course, the same. The only difference in the two cases is the horizontal component of F. Try again.

Hint: No need for any formulas with m or v !