Pretty please with a cherry on top?

[Imparcticle]

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-. :

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.

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.
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.

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(mv²)₁ - 1/2(mv²)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?

thank you!

 

[VietDao29]

Hi,
For the first question, assume that the car goes at a constant velocity. So F = 850 N.
How much work is done by the engine if the car goes for 17 km? (14.45 MJ)
So the engine will do 14.45 MJ using 1 litre of gasoline.
1 litre of gasoline can produce a 42 MJ work.
So the efficiency is $\frac{14.45}{42}$

Do the same for number 3. You must multiple by 100 then divide by 27.

For the second. You have already worked out the work done by gravity force (18228 J). So the work done by the cord must equal to the work done by gravity force to make that human have the same speed as he started jumped out of the bridge (i.e 0 m/s). So when $W_{rope} = W_{gravity}$ that human will stop going down and start going up.
$$W_{cord} = \frac{1}{2}k(x_{2}^{2} - x_{1}^{2})$$
Were x1 and x2 is the difference between the length of the cord at the time measured and the length of the normal cord (12m)
Note that when the man goes down 12m. The rope starts acting force. So x1 = 0m. And x2 = 31m - 12m = 19m.
I get arround 100 N/m (not 10N/m as you said).
Viet Dao,

 

[HallsofIvy]

VietDao29 has given good answers.

I would point out that your error in problem 2 was assuming that the 31-12= 19 m stretch was the "x" in "F= kx". If the 60 kg person were just hanging from the bungee cord at a distance 19 m then that would be true- but in this situation, going down to 19m causes the person to bound back upward. Yes, you should use conservation of energy to find k. The person's kinetic energy at the bottom is 0 so the change in potential energy falling 31 m must be accounted for in the word done by the bungee cord in stretching 19 m (notice the difference in distances!).

 

[Imparcticle]

Do the same for number 3. You must multiple by 100 then divide by 27.

number three:

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%?

W=1/2 (mv²)₂-1/2(mv¹)₂
= 1/2 (2220 kg * [80.0 km/h]²)₂ - 1/2 (2220 kg * [50 km/h]²)₁
=4329000 J= 4.329 MJ

[(4.329 MJ)100%]/27% =16.03 MJ = 1.603 x 10^3 MJ

For the second. You have already worked out the work done by gravity force (18228 J). So the work done by the cord must equal to the work done by gravity force to make that human have the same speed as he started jumped out of the bridge (i.e 0 m/s). So when that human will stop going down and start going up.

So the work done on the rope is 18228J.

Were x1 and x2 is the difference between the length of the cord at the time measured and the length of the normal cord (12m)
Note that when the man goes down 12m. The rope starts acting force. So x1 = 0m. And x2 = 31m - 12m = 19m.
I get arround 100 N/m (not 10N/m as you said).

I see. The rope's initial length is zero because that's where the initial measure ment is made.

W_rope=1/2 k (x₂² - x₁²)

working equation:
(2W)/(x²₂ - x²₁ = K

2 (18228 J) / ([19m]² - [0m]²) = 100.986 N/m = 1.0 x 10^2 N/m

wow! thank you so much!

So basically the elastic potential energy is going to equal the work done by gravity. THANX.