Assume a small 11 kg crate is attached to a parachute. The chute pulls up on the crate with 150 N.

a. Find Fnet

b. Find a
____ m/s²

c. Find v 5 seconds after release
_____ m/s

I'm really stuck on this. Thanks!

Respuesta :

Explanation:

a. The net force is the upward force of the chute minus the weight of the crate.

∑F = F − mg

∑F = 150 N − (11 kg) (9.8 m/s²)

∑F = 42.2 N

b. From Newton's second law, the net force equals the mass times acceleration:

∑F = ma

42.2 N = (11 kg) a

a = 3.84 m/s²

c. Acceleration is the change in velocity over change in time.  Assuming the crate is released from rest:

v = at + v₀

v = (3.84 m/s²) (5 s) + (0 m/s)

v = 19.2 m/s