Question

The position of a helicopter of weight=2.75 X 10^5N is r=(0.020m/s3)(t3)i + (2.2 m/s)(t)j - (0.060m/s2)(t2)k. What is the net force on the helicopter at t=5.0s?

Answer 1

F = 17170.16 N = 17.17 KN

Step-by-step explanation:

First we need to find the mass of helicopter by using its weight:

Weight = mg

2.75 x 10⁵ N = m(9.8 m/s²)

m = (2.75 x 10⁵ N)/(9.8 m/s²)

m = 28061.22 kg

Now, we find acceleration. We have position vector as:

r = (0.02 m/s³)(t³)i + (2.2 m/s)(t)j - (0.06 m/s²)(t²)k

taking its derivative twice, we can find acceleration:

a = (3)(2)(0.02 m/s³)(t)i + (0)j - (2)(1)(0.06 m/s²)k

a = (0.12 m/s³)(t)i - (0.12 m/s²)k

at, t = 5 sec

a = (0.12 m/s³)(5 s)i - (0.12 m/s²)k

a = (0.6 m/s²) i - (0.12 m/s²) k

Now, the magnitude of acceleration will be:

a = √[(0.6)² + (-0.12)²]

a = 0.61 m/s²

So, from Newton's Second Law, the net force on helicopter is given as:

F = ma

F = (28061.22 kg)(0.61 m/s²)

F = 17170.16 N = 17.17 KN