Question

A 4520 kg helicopter accelerates upward at 1.9 m/s2. What lift force is exerted by the air on the propellers?

Answer 1

`52884\,N`

Explanation:

Let 'm' denotes the mass of helicopter, 'g' denotes the gravity and 'a' denotes the acceleration.

We know that
`F_(grav)=mg`

Put
`m=4520 kg\,,\,g=9.8 \,\,m/sec^2`

So,

`F_(grav)=mg=4520(9.8)=44296\,N`

Also, as per Newton's second law,

`F_(motion)=ma`

Put
`m=4520 kg\,,\,a=1.9 \,\,m/sec^2`

So,

`F_(motion)=ma=4520(1.9)=8588\,N`

We know that the lift force is the sum of the gravitational force and the force of motion.

`F_(lift)=F_(grav)+F_(motion)\\=44296\,N+8588\,N\\=52884\,N`