Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

thushananth01 Group Title

A ball of mass 50g falls from a height of 2.0m and rebounds to a height of 1.2m. How much kinetic energy is lost on impact?

  • one year ago
  • one year ago

  • This Question is Closed
  1. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I need your help :) @Callisto @Vincent-Lyon.Fr @ganeshie8

    • one year ago
  2. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1354449116877:dw|

    • one year ago
  3. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact

    • one year ago
  4. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    At position 1: Total energy = PE = \(mgh_1\) At position 2: Total energy = (KE + energy loss) = total energy at position 1 So far so good?

    • one year ago
  5. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i got my answer as 0.5886 which is equal to PE after impact

    • one year ago
  6. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    (Ep+Ek) at 1 = (Ep + Ek) at 2 mgh = 1/2m(V,)^2 Kinetic energy at the beginning = mgh (Ep+Ek) at 2 = (Ep + Ek) at 3 mgh = mgh1 + 1/2m(v,,)^2 so Loss in kinetic energy = mgh1?

    • one year ago
  7. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmmm.. At position 3: Total energy = PE = \(mgh_2\) = KE at position 2 So, KE at position 2 = 0.05 x 9.81x 1.2 And from (KE + energy loss) = total energy at position 1 0.05 x 9.81x 1.2 + Energy loss = 0.05 x 9.81 x 2.0 So, you can get the energy loss

    • one year ago
  8. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    wht is the energy loss?

    • one year ago
  9. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1354449520295:dw|

    • one year ago
  10. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    What do you think?

    • one year ago
  11. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how is KE at position3 = 0?

    • one year ago
  12. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Velocity at the max point is 0, so KE=0 :|

    • one year ago
  13. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    note: in perfect bounce, the height of rebound is equal to the original height of drop. So, the loss in potential energy =loss in kinetic energy.(For quick answer)

    • one year ago
  14. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh... Is 1.2m the max. height of the rebound?

    • one year ago
  15. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    tht wsnt given in the question?

    • one year ago
  16. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Shadowys can u work it out using equations?

    • one year ago
  17. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    eh? Since it isn't a perfect rebound, loss of kinetic energy = loss of potential energy =\(mg(h-h_o)\) This is the potential energy is turned completely to kinetic energy during the drop, and vice versa during the rebound after hitting the ground.

    • one year ago
  18. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Shadowys what ws your answer? can u calculate and let me know

    • one year ago
  19. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @Callisto Kinetic energy at lowest point = Loss in potential energy 1/2mv^2 = mgh Ek = 0.05*9.81*2=0.981 Ep + Ek at 2 = Ep + Ek at 3 0.981 = 0.05*9.81*1.2 + Ek Ek at 3 = 0.3924J there loss in kinetic energy = K.E before impact - K.E after impact 0.981-0.3924= 0.5886

    • one year ago
  20. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    .392J

    • one year ago
  21. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    why would the Kinetic energy at lowest point = Loss in potential energy?

    • one year ago
  22. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy - energy loss due to impact

    • one year ago
  23. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek

    • one year ago
  24. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    mechanical energy is not really conserved since you've loss of rebound height, right?

    • one year ago
  25. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?

    • one year ago
  26. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.

    • one year ago
  27. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    one more question K.E at position3 = 0?

    • one year ago
  28. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.

    • one year ago
  29. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks:)

    • one year ago
  30. Shadowys Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you're welcome :)

    • one year ago
  31. Vincent-Lyon.Fr Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Final mechanical energy = Initial mechanical energy - energy loss. Hence Eloss = mg(h1-h2) = 0.40 J with g = 10 m/s²

    • one year ago
  32. thushananth01 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    correct!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.