An engine works at 26% efficiency. The engine raises a 5-kg crate from rest to a vertical height of 8 m, at which point the crate has a speed of 4 m/s. How much heat input is required for this engine?
2007-11-30 04:47:31 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Energy needed to lift = Gain in G.P.E + Gain in K.E
=(Mass X Grav acc. X height)+(1/2 X mass X velocity^2)
=(5X9.8X8)+(1/2X5X(4)^2)
=432 J.
Input = (100X432)/(26) = 1661.538462 J .
2007-11-30 05:02:59 · answer #1 · answered by Murtaza 6 · 0⤊ 0⤋
The work done by the engine appears as the changes in potential and kinetic energy of the crate.
W = mgh + 1/2 m v^2
Divide that by .26 to find the heat input for the engine.
2007-11-30 12:56:55 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋
The work done on the crate (at 5kg it's more of a box!) can be found by calculating the total energies of the crate.
It's got mgÎh of gravitational potential energy that's 5*9.8*8
It's got ½mv² of kinetic energy that's ½*5*4²
Add these two energies together to find the work done on the crate.
Once you've worked out the total work done on the crate, work out the heat requirements of the engine.
W=0.26H so H=W/0.26
OK?
2007-11-30 13:03:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋