The position of a 2.75×105n training helicopter under test is given by r⃗ =(0.020m/s3)t3i^+(2.2m/s)tj^−(0.060 m/s2)t2k^. part a find the net force on the helicopter at t=5.0s.

The net force on the helicopter at t=5.0s = F = 1.65.10⁴i - 3.3.10³k

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

F = force, N

m = mass = kg

a = acceleration due to gravity, m / s²

While the speed is the first derivative of distance  

While acceleration is the first derivative of the velocity function for t

It is known that the distance function (r) = (0.020 t³i + 2.2 tj-0.06t²k

Then the speed (v)

v = 0.06t²i + 2.2j - 0.12tk

Then acceleration (a)

a = 0.12ti - 0.12k

While :

W = m . g

m = 2.75.10⁵ : 10

m = 2.75.10⁴

so  net force :

F = m . a

F = 2.75.10⁴ . ( 0.12ti - 0.12k)

for t = 5  

F = 2.75.10⁴ . ( 0.12.5i - 0.12k)

F = 2.75.10⁴ (0.6i-0.12k)

F = 1.65.10⁴i - 3.3.10³k

Newton's second law

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the acceleration of an object

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a correct statement of Newton's second law

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Keywords : Newton's Law, acceleration, net force