An engine works at 23% efficiency. The engine raises a 7-kg crate from rest to a vertical height of 9 m, at which point the crate has a speed of 4 m/s. How much heat input is required for this engine?
2007-04-26 19:16:03 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The total energy change of the crate is m*g*h + 0.5*m*v^2. The amount of energy input needed to do this is (m*g*h + 0.5*m*v^2) / Eff, where Eff is the fractional efficiency (%Eff/100). This will come out in joules, which you can convert to calories.
(7kg*9.8m/s^2*9m + 0.5*7kg*4^2m^2/s^2) / 0.23,
E = 2.93*10^3 joule or 699.7 calories
2007-04-26 19:22:41 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
A little soft shoe
You increase both potential and kinetic energy of your crate. Figure out both.
Now convert your energy, which will (of course) be in newton-meter/sec to joules/sec.
Now divide that by 0.23.
2007-04-26 19:24:15 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋