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|Mar30-12, 07:45 AM||#1|
Find work given mass, acceleration, and time.
1. The problem statement, all variables and given/known data
A 510 kg, light-weight helicopter ascends from the ground with an acceleration of 2.30 m/s^2.
Over a 5.50 s interval, what is the work done by the lifting force?
2. Relevant equations
3. The attempt at a solution
Find force = m(a) = 510kg*2.3m/s^2 = 1173N
Find distance = .5(a)t^2 = 34.7875m
use this force and distance in w=f(d) to find work (in Joules):
1173*34.7875 = w
w = 40805.74J
|Mar30-12, 07:50 AM||#2|
Ok this is fairly simple, if the helicopter travels at a speed of 2.3m/s^2 and it traveled for 5.5 seconds, then how many meters did it travel in that 5.5 seconds? That is the distance.
|Mar30-12, 08:16 AM||#3|
I'm not going to ask you to do it for me, but using the equation above, I already solved for distance. Unless that is the wrong equation.
|acceleration, distance, mass, time, work|
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