Find work given mass, acceleration, and time.

[Gshaq Pierre]

Homework Statement

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

Homework Equations

f=m(a)
d=.5(a)t^2
W=F(Cosθ)*d

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.