How Does Angling the Force Affect Kinetic Energy Increase in a Sled?

[kaspis245]

Homework Statement

A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 38% . By what percentage would the sled's kinetic energy have increased if this force had pointed 62^o above the +x axis?

Homework Equations

The Work-Energy Theorem

The Attempt at a Solution

So I get, that the kinetic energy is reduced by 35% .

Correct answer: the kinetic energy increases by 18% .[/B]

 

[Bystander]

reduced

Correct answer: the kinetic energy increases

If the problem statement is complete and correctly transcribed in your original post, without checking your math, you are correct. It would be best that you double-check the problem statement.

 

[kaspis245]

The problem statement is correct 100%. Maybe I understood the problem incorrectly?

 

[Bystander]

As written, what you've done looks correct --- the x-component of the force is 100 % for the first case, and is reduced for the second. Looks like a really lousy problem statement, and an even worse job of proof-reading/checking matches of answers with problems. Let us know whether anyone fesses up, or how they rationalize the answer that's claimed.

 

[BvU]

I think it is understanding indeed. I fail to understand what W₀ represents, for example.
You are given that E_kin + F $\cdot$ s = 1.38 E_kin

Now try to grasp what you are being asked. The 17.8% book answer is correct.

 

[kaspis245]

W_o represents work with which other works (W₁ and W₂) are compared.

I understand your method, but I don't understand why my method didn't work.

 

[BvU]

Still don't understand. If you write W₁ = F $\cdot$ s , then I don't understand why you consider that to be 138 % of W₀ instead of 38%

 

[kaspis245]

As the problem says, the kinetic energy of W_o increases by 38% .

If I say, that the very first energy, with which we will compare other energies, is equal to 100% , then the second energy W₁, which is along +x axis, must be equal to 138% (increases by 38%) .

 

[BvU]

Fine with me. But then you must write W₁ = W₀ + F $\cdot$ s.
There is no friction, the original kinetic energy doesn't go away.

And energies are not along any axis. Energies are numbers. But I understand what you mean.

 

[Bystander]

BvU has you pointed in the right direction --- stay with him, and forget everything I said --- I misread the problem horribly.