A 2.0 kg mass starts from rest and slides down an inclined plane 8.0 X 10^-1 m long in .50s. what net force is acting on the mass along the incline.

Question

I have no idea how to do this

I don't know why they just didn't say .8m

the back of the book says 13 N down the incline

I can determine the mg with 9.81 m/s^2 and get 20 N but have no idea how to get the other forces

Answer 1

Vfinal = 2 Vaverage
= 2*0.8/0.5 = 3.2 m/s
a = Vfinal/time = 6.4 m/s^2

Answer 2

Fnet=?
a=6.4m/s^2
m=2.0 kg
Fnet=ma
Fnet=(2.0 kg)(6.4 m/s^2)
Fnet= 12.8 N
= 13 N

Answer 3

F=ma
F=2.0(9.81)
F=19.62
....?