We're doing test corrections for my physics class. I did really well, however there were three questions that I had no idea how I was supposed to solve. here are the pesky critters that landed me an A-. :grumpy: :(adsbygoogle = window.adsbygoogle || []).push({});

On this one, I guessed. I had and have absolutely no idea as to how in the world I'm supposed to do this. on the other two, I at least had some idea. 1.) What is the energy efficiency of a car which gets 17 km/litre if the car drives against friction and air resistance totalling 850N? The energy content of gasoline is 42 megajoules/litre.

so, can some one show me how to do this?

2.) A 60kg bungee jumper leaps from a bridge. She is tied to a 12 m long bungee cord and she falls 31m. Calculate the spring constant of the cord.

okay, on this one I thought I knew what to do. this is what I did:

The formula I need for this: F=kx, k is spring constant and x is the stretch and F is the force exerted. So all I have to do is solve for k....but I need force. To find force I naturally calculated the force from the bungee jumper by F=ma and got 588 N.

If she's tied to a 12m long bungee cord, and it stretched to 31m, I figured I'd subtract 31-12 and get 19. This is the value for x.

my working equation is then F/x=k. By plugging in values, I got 3.1 x 10^1 N/m for k. however, this was the incorrect answer.

so when I was trying to correct the answer, i figured that I probably should have calculated the work done on the bungee cord, so W=Fd=588N * 31m= 18228 J. my working equation is still F/x=k. so by plugging in the appropriate values, my answer was about 959 N/m or 9.6 x 10^2 N/m. Am I doing my significant figures incorrectly or something? The answer the teacher gave was 1.0 x 10^1 N/m, and that's pretty close.

anyway, next problem.thank you!! how much work is required to accelerate a car with mass 2220 kg from 50.0 km/h to 80km/h in 10sec if its engine's efficiency is 27%?

my mind goes blank at the sight of "efficiency". my guess as to how this problem is done: The first part would be done by the fact that W=deltaKE or Work=1/2(mv^{2})_{1}- 1/2(mv^{2})_{2}. The work you get here is with 100% efficiency so input/output=100% work/27% of the work. IOW, you'd find 27% of the work, and divide it into 100% of the work. Am I at least close??????????

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# Homework Help: Pretty please with a cherry on top?

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