Find work given mass, acceleration, and time.

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SUMMARY

The discussion focuses on calculating the work done by a helicopter with a mass of 510 kg ascending with an acceleration of 2.30 m/s² over a time interval of 5.50 seconds. The force exerted by the helicopter is calculated using the formula F = m(a), resulting in 1173 N. The distance traveled is determined using d = 0.5(a)t², yielding approximately 34.79 m. The total work done is then calculated as W = F * d, resulting in 40,805.74 Joules.

PREREQUISITES
  • Understanding of Newton's Second Law (F = m(a))
  • Knowledge of kinematic equations for uniformly accelerated motion
  • Familiarity with the work-energy principle (W = F * d)
  • Basic proficiency in unit conversions (e.g., kg to N, m to Joules)
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their teaching of work and energy concepts.

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Gshaq Pierre
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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
 
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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.
 
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.
 

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