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I need your help :) @Callisto @Vincent-Lyon.Fr @ganeshie8
yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact
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?
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
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.
you're welcome :)
Final mechanical energy = Initial mechanical energy - energy loss. Hence Eloss = mg(h1-h2) = 0.40 J with g = 10 m/s²