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The position of a 2.75X10^5 N training helicopter under test is given by

r=(.020m/s^3)t^3i(i)+(2.2m/s)t(j)-(.060m/s^2)t^2(k)

Find the net force on the helicopter at t=5s

my answer: 6(0.020)5(i)-2(0.060)(k)

response:Your answer either contains an incorrect numerical multiplier or is missing one.?

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Force is mass times acceleration.

Just looking at your first term (I can't read the last term for "r") you have calculated the acceleration by twice differentiating "r" , then evaluated it at t=5s. That's all fine for acceleration but now you have to multiply each term by the mass, (2.75/9.8)x10^5 kg

So, assuming both terms in your acceleration are correct, the force is then;

F = (2.75/9.8)x10^5x6x.02x5i - (2.75/9.8)x10^5x2x.06k

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