What Went Wrong with my 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 !

 

[M98Ranger]

First of all, it's great that you provided your calculations and showed your thought process. It's important to double check your work when solving problems like this, so let's take a closer look at your calculations.

In your first case, where the force is parallel to the ground, you correctly used the formula for work to determine the change in kinetic energy. However, in your second case where the force is at an angle of 38°, you used the same formula but with a different value for the work done (100 x y). This is where the error lies.

The work done in both cases should be the same, since the only difference is the angle of the force. So, in your second case, the work done should also be 100 x 0.47. This would give you a value of y=0.47, resulting in a 47% increase in kinetic energy.

Overall, your thought process was correct, but there was a small error in your calculations. Keep in mind to double check your work and make sure all the variables and formulas are consistent. Good luck with your future problem solving!