Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

I need your help :) @Callisto @Vincent-Lyon.Fr @ganeshie8
|dw:1354449116877:dw|
yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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?
yes i got my answer as 0.5886 which is equal to PE after impact
(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?
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
wht is the energy loss?
|dw:1354449520295:dw|
What do you think?
how is KE at position3 = 0?
Velocity at the max point is 0, so KE=0 :|
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)
Oh... Is 1.2m the max. height of the rebound?
tht wsnt given in the question?
@Shadowys can u work it out using equations?
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.
@Shadowys what ws your answer? can u calculate and let me know
@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
.392J
why would the Kinetic energy at lowest point = Loss in potential energy?
Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy - energy loss due to impact
Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek
mechanical energy is not really conserved since you've loss of rebound height, right?
ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?
yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.
one more question K.E at position3 = 0?
yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.
thanks:)
you're welcome :)
Final mechanical energy = Initial mechanical energy - energy loss. Hence Eloss = mg(h1-h2) = 0.40 J with g = 10 m/s²
correct!

Not the answer you are looking for?

Search for more explanations.

Ask your own question