Calculating Percent Change in Kinetic Energy with Varying Force Angles

[leezak]

% change in KE

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 36.7%. By what percentage would the sled's kinetic energy have increased if this force had pointed 62.2° above the +x axis?

i'm not sure how to even begin this problem... I'm guessing i should use the change in KE = Work but I'm not sure how to go about it. help? thanks!

 

[Päällikkö]

You are correct: W = Fx.
We can assume that only the horizontal component of force increases the kinetic energy.

What is the horizontal component of F in case one (where F is horizontal)? How about in case two (where F is tilted)?

Can you figure out how to get an equation describing the situation (increase in kinetic energy) in case one? How about in case two?

 

[leezak]

if X is the horizontal component then x*cos(62.2) is the hypotenuse for the second case, right? I'm not sure how to make an equation for that. i know that PE at the bottom of the hill is 0 and therefore it must have a higher KE at the bottom and has a higher PE at the top. I'm really not sure how to make an equation out of that though

 

[Päällikkö]

Hill? What hill? The hypotenuse is F. What is F's horizontal component?

 

[leezak]

F*cos(62.2) is F's horizontal component... how can i use that in an equation?

 

[Päällikkö]

Only the horizontal F increases kinetic energy, unless the sled takes off (as in: airplanes do), but we assumed it doesn't.

Anyways, this is the situation:
K₁ + W = K₂
In the beginning, the sled has some kinetic energy (K₁). Some (positive) work is done to the sled (W) and its kinetic energy increases (K₂).

Can you use the information given in the problem to get two equations (case one and case two)?

 

[leezak]

does this make sense... for situation one --> (1/2)*m*(v^2) + (f*d) = .367 * (1/2)*m*(v^2) and for situation two --> (1/2)*m*(v^2) + (f*d*cos(62.2)) = (1/2)*m*(v^2) ? how can i use those

 

[Päällikkö]

Close, but needs some adjusting:
(1/2)*m*(v^2) + (f*d) = .367 * (1/2)*m*(v^2)
Multiplying by 0,367 would mean the work was negative (as the kinetic energy decreases). What should you multiply the final kinetic energy with?

(1/2)*m*(v^2) + (f*d*cos(62.2)) = (1/2)*m*(v^2)
You must distinguish the energy in the beginning and in the end. Your equation gives fdcos(62,2) = 0, which is not the case. As you will be wanting the ratio between energy in the beginning and the end, I suggest multiplying the final energy by some A (which would be related to the asked percentage).

As you are not asked for velocity, using (1/2)mv^2 is unnecessary. Use K.

 

[leezak]

instead of multplying by .367 should i divide by .367?

 

[Päällikkö]

As there's an 36,7 % increase: E_initial + 0,367E_initial = E_final

So, you should multiply it by 1,367.

 

[leezak]

so now i have the two equations...
K1 + f*d = 1.367*K2 and
K1 + f*d*(cos(62.2)) = A*K2
but i have a lot of variables so what do i solve for?

 

[Päällikkö]

K2 is actually K1.
A is the asked quantity. Solve for it.

Well, now it's just mathematics. Give it a shot, ask for more help if you can't get it solved .

 

[leezak]

what about fd though

aren't there still 3 variables... fd, A, and K1

 

[Päällikkö]

fd will cancel out.

 

[leezak]

from equation one i got K1 = fd/.367 and i plugged that into equation two and eventually got A=1.46... I then multiplied that by 100 to get 146%, but that's wrong... I'm not sure what i did wrong

 

[Päällikkö]

Ok, K = fd / .367.
K + fdcos(62.2) = AK
fd / .367 + fdcos(62.2) = A(fd / .367)
fd(1/.367 + cos(62.2)) = fdA / .367 _____[/color] | fd cancels out
A = 1 + .367cos(62.2) = 1.1712

=> 17.1 %
I hope that's correct .

 

[leezak]

oh so you have to take out that one before putting into a percentage... thanks!