Consider a frictionless slide inside a city park. A 49kg child sits on the top of a slide that is located 1.80m above the ground. After her descent, the child reaches a velocity of 3.00m/s at the bottom of the slide. Find value for % efficiency.
I missed class and so I don't know how to do it. It's probably easy but I missed the lesson. Help please.
2007-04-17 11:27:35 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
work (energy) to get her up the ladder:
W = Force * distance
W = (mass * acceleration) * 1.8 m
W = 49 kg * 9.8 m/s^2 * 1.8 m
W = 864.36 Joules
energy off the slide:
E = 1/2 * m * v^2
E = 1/2 * 49 * 3.00^2
E = 220.5 Joules
efficiency = work out / work in
eff. = 220.5 / 864.36 = 25.5%
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2007-04-17 11:34:34 · answer #1 · answered by tlbs101 7 · 0⤊ 0⤋
If the child fell 1.8 metres to the ground with no friction or other energy losses, she'd have kinetic energy mgh=49x10x1.8=882 joules. With a velocity 3m/s at the bottom of the slide she has k.e. 1/2 mv^2=.5x49x9=220.5 joules; she's lost the rest of the 882 joules to friction, so percentage efficiency=220.5/882x100=25%.
2007-04-17 18:40:17 · answer #2 · answered by zee_prime 6 · 0⤊ 0⤋
29.4% someone told me thats the exact ancient computation secret''while 25% is just minimal
2007-04-17 18:39:14 · answer #3 · answered by etanmi4 2 · 0⤊ 0⤋