Find work given mass, acceleration, and time.

Gshaq Pierre

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?

a=2.3m/s^2
m=510kg
t=5.5s




2. Relevant equations


f=m(a)
d=.5(a)t^2
W=F(Cosθ)*d



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

mtayab1994

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.

Gshaq Pierre

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.