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Work and Energy Problems

  1. Jan 8, 2008 #1
    [SOLVED] Work and Energy Problems

    I have a few work and energy problems I don't understand at all. Please help.

    While traveling to school at 27 m/s your car runs out of gas 15 km from the nearest gas station. If the station is 16 m above your current elevation, how fast will the car be going when it reaches the gas station? Ignore friction. Answer 20.4 m/s. Problems must solved using energy principles.

    There's only kinetic energy so I would use K = 1/2 mv^2. After that, I don't know what to do.

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    A lead ball is traveling at 275 m/s when it strikes a stell plate and comes to a stop. If all its kinetic energy is converted into heat and none of the energy leaves the bullet, what is the bullet's change in temperature? (C = 125 J/ kg * k). Answer: 302.5 degrees C.

    I tried using the equation Q = mcT but there no information on either initial or final temperature so I don't know what to do. I tried setting U = K but height's not given.

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    A 1.5 kg bass is hooked by a fisherman. The fisherman plays the bass by allowing the fish to swim off 2.1 m/s before braking his reel and stopping the bass in 37 cm. How much tension is exerted on the line? Assume the fish is neutrally buoyant. Answer: 8.94 N

    I have no clue on this problem. All I know is that I have to use k = 1/2 mv^2 because it involves velocity.
     
    Last edited: Jan 9, 2008
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  3. Jan 8, 2008 #2

    Dick

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    No offense, but those are really poor attempts. Take the first one, how about using potential energy=mgh? The second one is about CHANGE in temperature vs CHANGE in heat which you can deduce from the kinetic energy. Not about absolute temperature. And for the last one, how about work=force times distance? Please try again.
     
  4. Jan 8, 2008 #3
    I did try harder but I just didn't include it all. I already got to the part where I know K final = K initial - U final but the problem is that I don't know what do with the 15 km.
     
  5. Jan 8, 2008 #4

    Dick

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    Do nothing with it. It doesn't matter. Sometimes problems include irrelevant information. You know that, right?
     
  6. Jan 9, 2008 #5
    in the first question, the 15km means nothing because you can ignore friction. your equation is correct, K final = K initial - (U final - U initial), remember, mass cancels out of the equation.

    second equation, you have the total energy of the system in the bullets velocity, you know that it takes 125 J to change the temperature of the bullet 1 K. see if that helps you.

    third, work = force*distance. start there, along with the 1/2mv^2 equation. demintional analysis will help on all of these problems as well.
     
  7. Jan 9, 2008 #6
    For the second problem, I don't know what to do with the velocity because mass is not given so I can't use K = 1/2 mv^2. You're not given height either so there's no way to find the potential energy.

    For the third problem, I use W = K and so I add K with work = force*distance. The answer is slightly off.
     
    Last edited: Jan 9, 2008
  8. Jan 9, 2008 #7

    hage567

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    For question #2 you're looking for the change in temperature. You just solve for deltaT. I don't know why you are trying to bring height into things.

    For question #3, show your calculation.
     
  9. Jan 9, 2008 #8
    I'm sooo confused. For 2, you're not given mass and you can't set K = 1/2 mv^2 equal to Q = mcT. You can't cancel it out even if you used W = K because Q is not work. I need help getting v to work.

    For 3, I did 1/2(1.5)(2.1^2) + 1.5(9.8)(.37) = 8.47 J
    I did 1/2mv^2 + Fx.
    Edit: I just realized this is work instead of force of tension...
     
    Last edited: Jan 9, 2008
  10. Jan 9, 2008 #9

    hage567

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    Notice, that the mass term is on both sides of your equation. It therefore cancels out and you don't need to know what it is. Q is energy, and 1/2mv^2 is energy, so yes you can set them equal since all of the kinetic energy is going into heating the bullet.
     
  11. Jan 9, 2008 #10
    I finally got it to work. I never knew you could set K equal to Q.

    How can I get work to be force of tension?
     
    Last edited: Jan 9, 2008
  12. Jan 9, 2008 #11

    hage567

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    Why is there a 9.8 in there? This doesn't have anything to do with gravitational potential energy. The tension is the force applied to stop the fish. The work done in stopping the fish is this force times the distance over which it acts to bring the fish to rest.
     
  13. Jan 9, 2008 #12
    Thank you so much. I finally figured it out. I would use W = K and set that to W = Fxcos(theta) Then I divide K by x which gives me F which is the force of tension.
     
  14. Jan 9, 2008 #13

    hage567

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    You're welcome. :smile:
     
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