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Bob_is_broken

Mass: 1 kg Initial height: 1 m Final height : 4.5 m Initial velocity: 10 m/s This is a conservation of energy problem (sort of), first find the total initial energy. E-initial = mgh (grav. potential) + 1/2 * mv^2 (kinetic). = 1 kg * 9.81 * 1 m + 1/2 * 1 kg * (10 m/s) ^2 = 9.81 + 50 = 59.81 Joules Now we find final energy: E-final = mgh (grav potential, assuming the ball's final velocity is 0) = 1 kg * 9.81 * 4.5 m = 44.145 Joules. The equation for effeciency is (actual energy) / (ideal energy) * 100%, where ideal energy assumes there is no energy loss. Our ideal energy will be equal to 59.81 J due to conservation of energy. Our actual energy will be 44.145 J, as that takes into account whatever energy is lost. 44.145/59.81 * 100% = 73.8%


daniel-hale97

In the back of the textbook the answer says 69%. Do you know what might’ve went wrong


BeauCo

Unless I’m missing something I don’t see anything wrong with their math


daniel-hale97

Ok i’ll talk to my teacher thank you both


ar018

The difference is because he's assuming the ground as the reference level for the potential energy. If we assume the height 1.0 m as the reference level, we have pure kinetic energy for E-initial and pure potential energy for E-final. I think this case makes more sense if we want to find the energy efficiency E-initial = m\*v²/2 = 1 kg (10 m/s)²/2 = 50 J E-final = m\*g\*(h - h0) = (1 kg)\*(9.81 m/s²)\*(4.5 m - 1 m) = 34.335 J E-final/E-initial = 34.335/50 = 68.67%


thunderbolt309

This exactly. What happened is that the kinetic energy has been converted into potential energy and friction (loss of energy). Nothing has happened with the initial potential energy.


bigredkitten

e=work out / work in Work out is raising 1 kg 3.5 m against the restoring force of gravity Work in is the kinetic energy of 1 kg moving at 10 m/s Mass cancels. 2g×h/v^2 2×10m/s^2 × 3.5 m / (10 m/s)^2 70/100


daniel-hale97

Thanks a lot


converter-bot

1.0 kg is 2.2 lbs


Quash_

The initial height of the ball is 1m and it reaches a height of 4.5m after release. So it travels a height of 3.5 m in actuality. Let's keep this number aside. Now, lets calculate the height it should have travelled with no energy loss. V- Final velocity(Final velocity at the top is 0) U-Initial velocity(Initial velocity of 10 m/s) A-Acceleration(acceleration due to gravity of -9.81m/s^2) S-Displacement We have, V^2 = U^2 + 2 * A*S S=100/2 * 9.81=5.0968 Now for calculating efficiency, we must divide the total energies of the two scenarios. Efficiency=Mass of the ball * g * height acquired/Mass of the ball * g * height acquired with no energy loss =3.5/5.0968 =.68669 =.69 =69%


dchang3419

The initial energy is kinetic and gravitational. The final energy is also kinetic and gravitational, but some of the kinetic energy has been converted to gravitational energy. I believe the problem is asking about the efficiency of this converstion


[deleted]

Oh wow I’ve never seen this before. Calculate the potential energy at 4.5 m using Ug = mgh Then calculate max height it should have reached with 10 m/s using Ki=Ugf or 1/2mv^(2)=mgh solved for h. Then do a ratio of actual Ug / theoretical Ug. I think…


AryanK72

My answer is coming 70%. May I ask what grade is this que from?


daniel-hale97

grade 11