Does anyone know how to answer this?

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Fliw42
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#1
Report Thread starter 2 years ago
#1
The assumption that there is no energy loss is not accurate. Assuming the spring has a spring constant of 100Nm-1 and has a length change of 2 cm, the track is on a level surface, and that the final velocity of the car is 0.25 ms-1 when it reaches the end of the track, what is the percentage efficiency of the toy? Take the mass of the car to be 150 g

The answer = 23% but I don't know how to get there
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David Getling
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#2
Report 2 years ago
#2
(Original post by Fliw42)
The assumption that there is no energy loss is not accurate. Assuming the spring has a spring constant of 100Nm-1 and has a length change of 2 cm, the track is on a level surface, and that the final velocity of the car is 0.25 ms-1 when it reaches the end of the track, what is the percentage efficiency of the toy? Take the mass of the car to be 150 g

The answer = 23% but I don't know how to get there
You have only given part of the question, but I think what's wanted is the final KE of the car divided by the initial PE stored in the spring.
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lzaak
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#3
Report 2 years ago
#3
(Original post by Fliw42)
The assumption that there is no energy loss is not accurate. Assuming the spring has a spring constant of 100Nm-1 and has a length change of 2 cm, the track is on a level surface, and that the final velocity of the car is 0.25 ms-1 when it reaches the end of the track, what is the percentage efficiency of the toy? Take the mass of the car to be 150 g

The answer = 23% but I don't know how to get there
seneca work

you need equations elastic potential energy and kinetic energy

work out both

ke equation is 1/2mv^2
m = 150
v = 0.25 (ms^-1 = m/s therefore no change)
therefore 0.5*150*0.25 = 4.6875

epe equation is 1/2ke^2

k = 100^-1 = 10 (has to be changed because unit is Nm not Nm^-1)
e = 2^2 = 4
so the answer would be 0.5*4*10 = 20

In the replies david is right about how the question was cut in half but to know which way you divide it we need how the energy is converted, but if you don't know still use logic, so to find the efficiency you do (4.6875* 100) / 20 equal around 23%
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Asheri2006
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#4
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#4
wait i dont get it, why did you change the 100^-1 but no the 0.25ms^-1?
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Eimmanuel
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#5
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#5
(Original post by Asheri2006)
wait i dont get it, why did you change the 100^-1 but no the 0.25ms^-1?
I would suggest that you work it out yourself.
You just need to substitute in the given values into the KE formula and elastic PE formula to obtain the answer.
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Asheri2006
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#6
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#6
(Original post by Eimmanuel)
I would suggest that you work it out yourself.
You just need to substitute in the given values into the KE formula and elastic PE formula to obtain the answer.
umm that didn't have much to do with it but okayy
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Eimmanuel
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#7
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#7
There is no point in just following what Izaak had done.

(Original post by Asheri2006)
wait i dont get it, why did you change the 100^-1 but no the 0.25ms^-1?
IMO, Izaak was doing something inconsistent and wrongly.

The unit of spring constant is N/m NOT Nm.

Next the following is also incorrect.
ke equation is 1/2mv^2
m = 150
v = 0.25 (ms^-1 = m/s therefore no change)
therefore 0.5*150*0.25 = 4.6875
https://www.wolframalpha.com/input/?i=0.5*150*0.25%5E2.

The unit of KE is NOT joule with this calculation but is some other unit of energy.

k = 100^-1 = 10 (has to be changed because unit is Nm not Nm^-1)
e = 2^2 = 4
so the answer would be 0.5*4*10 = 20
This calculation also does not give the unit of joule.

(Original post by Asheri2006)
umm that didn't have much to do with it but okayy
It has a lot to do with that. As the calculations are inconsistent, it is better not to follow them and I am asking you to do the calculations yourself.
This is a simple question and the only "difficulty" is already highlighted in post #2.
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theorigamicrane
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#8
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#8
Hey, here's how you work it out.

The efficiency equation is ( useful output energy / total input energy ) * 100 (for a percentage). It's not direct but I think the "total input energy" would be the EPE and the "useful output energy" would be the amount that was transferred to the toy car's kinetic energy store. You have to work out both.

You have to be careful with units here.

The equation for EPE = 1/2 * k (N/m) * e^2 (m)

k = 100 N/m

You're given e (extension) in cm! You have to convert this to metres to get the right answer, so that's 0.02 m

So you do 1/2 * 100 * 0.02^2 = 0.02 J

Now for kinetic energy:

KE = 1/2 * mass (kg) * velocity^2 (m/s)

You're given the mass as 150g, in kilograms, this would be 0.15 kg.

v = 0.25 m/s

So 1/2 * 0.15 * 0.25^2 = 4.6875 * 10^-3


Now you have both your "useful" and "total inputted" energy, you can work out the efficiency of the toy car.

( 4.6875 * 10^-3 / 0.02 ) * 100 = 23.4 % (3 s.f)
8
The Butterfly
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#9
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#9
S-o basically al u have to do is to calculate he Ee and the Ek and the efficiency Ee = 1/2 * 100 * 0.02 m squared = 0.02 EK = 1/2 * 0.15 kg * 0.25 squared = 0.0046875After that u just calculate the efficiency 0.0046875/ 0.02 * 100% = 23 % I think this should be right, my friend helped me with this ! I hope it helps
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emily_y_chen
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#10
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#10
(Original post by theorigamicrane)
Hey, here's how you work it out.

The efficiency equation is ( useful output energy / total input energy ) * 100 (for a percentage). It's not direct but I think the "total input energy" would be the EPE and the "useful output energy" would be the amount that was transferred to the toy car's kinetic energy store. You have to work out both.

You have to be careful with units here.

The equation for EPE = 1/2 * k (N/m) * e^2 (m)

k = 100 N/m

You're given e (extension) in cm! You have to convert this to metres to get the right answer, so that's 0.02 m

So you do 1/2 * 100 * 0.02^2 = 0.02 J

Now for kinetic energy:

KE = 1/2 * mass (kg) * velocity^2 (m/s)

You're given the mass as 150g, in kilograms, this would be 0.15 kg.

v = 0.25 m/s

So 1/2 * 0.15 * 0.25^2 = 4.6875 * 10^-3


Now you have both your "useful" and "total inputted" energy, you can work out the efficiency of the toy car.made

( 4.6875 * 10^-3 / 0.02 ) * 100 = 23.4 % (3 s.f)
made an account to like this lol
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