A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
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?
 2 years ago
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?

This Question is Closed

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0I need your help :) @Callisto @VincentLyon.Fr @ganeshie8

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1At position 1: Total energy = PE = \(mgh_1\) At position 2: Total energy = (KE + energy loss) = total energy at position 1 So far so good?

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0yes i got my answer as 0.5886 which is equal to PE after impact

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.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?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Hmmm.. 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

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0wht is the energy loss?

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0how is KE at position3 = 0?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Velocity at the max point is 0, so KE=0 :

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1note: 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)

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Oh... Is 1.2m the max. height of the rebound?

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0tht wsnt given in the question?

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0@Shadowys can u work it out using equations?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1eh? Since it isn't a perfect rebound, loss of kinetic energy = loss of potential energy =\(mg(hh_o)\) This is the potential energy is turned completely to kinetic energy during the drop, and vice versa during the rebound after hitting the ground.

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0@Shadowys what ws your answer? can u calculate and let me know

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.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 0.9810.3924= 0.5886

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1why would the Kinetic energy at lowest point = Loss in potential energy?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.1Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy  energy loss due to impact

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1mechanical energy is not really conserved since you've loss of rebound height, right?

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0one more question K.E at position3 = 0?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.0Final mechanical energy = Initial mechanical energy  energy loss. Hence Eloss = mg(h1h2) = 0.40 J with g = 10 m/s²
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.