Some questions related to work,energy and momentum

[shemer77]

Homework Statement

1)A ball of mass 1.71 kg is dropped from a height y1 = 1.38 m and then bounces back up to a height of y2 = 0.88 m. How much mechanical energy is lost in the bounce? The effect of air resistance has been experimentally found to be negligible in this case, and you can ignore it.

What I did was I used the energy before = energy after. And I got 6.161J, however it seems that is wrong...

2)A car of mass 917 kg is traveling on a horizontal segment of a freeway with a speed of 58.8 mph. Suddenly, the driver has to hit the brakes hard to try to avoid an accident up ahead. The car does not have an ABS (antilock braking system), and the wheels lock, causing the car to slide some distance before it is brought to a stop by the friction force between the car's tires and the road surface. The coefficient of kinetic friction is 0.301. How much mechanical energy is lost to heat in this process?
Im not sure even how to start up this problem, I've never done anything with heat :(

3)During an ice-skating extravaganza, Robin Hood on Ice, a 51.0-kg archer is standing still on ice skates. Assume that the friction between the ice skates and the ice is negligible. The archer shoots a 0.105-kg arrow horizontally at a speed of 99.4 m/s. At what speed does the archer recoil?
What I did was that there should be no change in momentum, but every time I solve it the computer says I have the wrong answer?? isn't it 0 = 51*v - .105*99.4 so v = .204 m/s?

4)A bungee jumper with mass 58.5 kg reaches a speed of 14.2 m/s moving straight down when the elastic cord tied to her feet starts pulling her back up. After 0.0250 s, the jumper is heading back up at a speed of 11.4 m/s. What is the average force that the bungee cord exerts on the jumper? and what is the average number g's experienced?
for the first part of that question I caculated the change in momentum which is impulse and I divided that by time and shouldn't that come out to the average force? I got 6552N, but apparently that is wrong, and how do you caculate the g's experienced?

I hate my teacher, and I AM TRYING TO LEARN, so please just don't give me the answer.

 

[rock.freak667]

Homework Statement

1)A ball of mass 1.71 kg is dropped from a height y1 = 1.38 m and then bounces back up to a height of y2 = 0.88 m. How much mechanical energy is lost in the bounce? The effect of air resistance has been experimentally found to be negligible in this case, and you can ignore it.

What I did was I used the energy before = energy after. And I got 6.161J, however it seems that is wrong...

I think you mean you found the change in energy. But you should check over your numbers. I believe you should get around 8 not 6.

2)A car of mass 917 kg is traveling on a horizontal segment of a freeway with a speed of 58.8 mph. Suddenly, the driver has to hit the brakes hard to try to avoid an accident up ahead. The car does not have an ABS (antilock braking system), and the wheels lock, causing the car to slide some distance before it is brought to a stop by the friction force between the car's tires and the road surface. The coefficient of kinetic friction is 0.301. How much mechanical energy is lost to heat in this process?
Im not sure even how to start up this problem, I've never done anything with heat :(

Worry not! What they're asking you is how much energy is lost due to friction basically. So the frictional force must bring the car to 0 mph, so the energy lost due to friction is the same as the change in?

3)During an ice-skating extravaganza, Robin Hood on Ice, a 51.0-kg archer is standing still on ice skates. Assume that the friction between the ice skates and the ice is negligible. The archer shoots a 0.105-kg arrow horizontally at a speed of 99.4 m/s. At what speed does the archer recoil?
What I did was that there should be no change in momentum, but every time I solve it the computer says I have the wrong answer?? isn't it 0 = 51*v - .105*99.4 so v = .204 m/s?

Yes there should be no change in momentum, but his mass is not 51 kg, remember that he is holding the arrow as well.

4)A bungee jumper with mass 58.5 kg reaches a speed of 14.2 m/s moving straight down when the elastic cord tied to her feet starts pulling her back up. After 0.0250 s, the jumper is heading back up at a speed of 11.4 m/s. What is the average force that the bungee cord exerts on the jumper? and what is the average number g's experienced?
for the first part of that question I caculated the change in momentum which is impulse and I divided that by time and shouldn't that come out to the average force? I got 6552N, but apparently that is wrong, and how do you caculate the g's experienced?

Can you show the working here? You might be doing it correctly but messing up the signs.

 

[shemer77]

I think you mean you found the change in energy. But you should check over your numbers. I believe you should get around 8 not 6.

Well now that I think about it isn't it 1/2mv2+mgh=1/2mv^2+mgh. and at both times the ke is zero so its just mgh=mgh. So the change would be mgh(f)-mgh(i)?

Worry not! What they're asking you is how much energy is lost due to friction basically. So the frictional force must bring the car to 0 mph, so the energy lost due to friction is the same as the change in?

i don't really understand... is it the change in momentum?

Yes there should be no change in momentum, but his mass is not 51 kg, remember that he is holding the arrow as well.

but that shouldn't make any difference though, does it? because momentum intial = momentum final. and his intial momentum is zero because his velocity is zero and afterward it would be 51*v - .105*99.4. I know I'm wrong but i don't understand. his final momentum mass would not include the mass of the arrow

Can you show the working here? You might be doing it correctly but messing up the signs.

i got this one, it was 59876 N, but how do i calculate the g's now?

 

[rock.freak667]

Well now that I think about it isn't it 1/2mv2+mgh=1/2mv^2+mgh. and at both times the ke is zero so its just mgh=mgh. So the change would be mgh(f)-mgh(i)?

Yes.

i don't really understand... is it the change in momentum?

What type energy does the car have and must be converted into heat so that it stops?

but that shouldn't make any difference though, does it? because momentum intial = momentum final. and his intial momentum is zero because his velocity is zero and afterward it would be 51*v - .105*99.4. I know I'm wrong but i don't understand. his final momentum mass would not include the mass of the arrow

Yeah, sorry I was thinking of something else. The method is correct, but try v = -0.204 m/s.

i got this one, it was 59876 N, but how do i calculate the g's now?

Well 1 g = 9.81 m/s² so just divide the 59876 by 9.81 and the force will be in g's.

 

[shemer77]

Yes.




What type energy does the car have and must be converted into heat so that it stops?



Yeah, sorry I was thinking of something else. The method is correct, but try v = -0.204 m/s.



Well 1 g = 9.81 m/s² so just divide the 59876 by 9.81 and the force will be in g's.

1) well i did that and i got -8.37 J but the computer says that is wrong :/
2)it has kinetic energy?
3)I tried -.204 m/s still wrong :(
4)so your saying the person experienced 6103 g's?

 

[rock.freak667]

1) well i did that and i got -8.37 J but the computer says that is wrong :/
2)it has kinetic energy?
3)I tried -.204 m/s still wrong :(
4)so your saying the person experienced 6103 g's?

1.You would need to input 8.37 J I would think.

2. So get the change in KE.

3. Not sure, what you're doing should be correct.

4. Yes that is a large number, but the jumper when from 14 m/s downwards to 11 m/s upwards in 0.025 seconds

 

[vela]

Well now that I think about it isn't it 1/2mv2+mgh=1/2mv^2+mgh.

You can't equate the initial energy with the final energy because energy isn't conserved when the ball bounces.

and at both times the ke is zero so its just mgh=mgh.

This tells you is h_i=h_f. In other words, if energy were conserved, the ball would bounce back up to the same height.

So the change would be mgh(f)-mgh(i)?

What you should have said was the initial and final energies were
E_i = \frac{1}{2}mv_i^2 + mgh_i = mgh_i
and
E_f = \frac{1}{2}mv_f^2 + mgh_f = mgh_f
But don't set them equal because that would mean energy is conserved, which is isn't in this case. The change in energy would still be \Delta E = mg(h_f - h_i).

I'm wondering if the computer is rejecting your answer because of rounding. I found the difference was -8.39 J, not -8.37 J. Generally, it's a good idea to keep extra digits on intermediate results and only round off to the correct number of significant figures at the very end. Also, as rock.freak noted, the question asks how much energy was lost, so you want to give a positive answer.

With the ice skater problem, again, it might be rounding. I found the archer's speed was 0.2046 m/s, which rounds to 0.205 m/s.

Remember that g is an acceleration, so you want to calculate the acceleration the jumper experienced. You calculated the force the bungee cord exerted. Once you have the acceleration in m/s², you divide it by 9.81 m/s² to find the number of g's the jumper felt. Given the numbers in the problem, you'll still get an unrealistic answer. As rock.freak noted, 25 ms is a ridiculously short time. It sounds like it's off by a factor of 100 or so.

 

[shemer77]

ah ok thanks vela and rock.freak, one last question, for the car one i think i understand it as all the kinetic energy has been converted to friction heat, so that's why the change in kinetic energy is equal to the heat loss. However am I correct in saying that the change in kinetic energy is 1585236.24 J?

 

[vela]

It's better to say the kinetic energy is converted by friction into heat. Also, it doesn't make sense to say "the change in kinetic energy is equal to the heat loss" because there's no heat loss in this problem. If anything, the tires and the road gain heat. Sorry if it seems I'm being picky, but I've found that this kind of word salad usually indicates a lack of understanding. It's important to learn to express yourself accurately, rather than relying on the reader to figure out what you really meant.

And, finally, to answer your question: No, that's not the correct change in kinetic energy.

 

[shemer77]

ah ok and thanks, i forgot to convert mph to m/s