Assuming "bottom of the incline" means the mass is on the incline already, with a velocity of 5.0 m/s parallel to the incline surface.
Fa - Ff - Fg = 0
(3 kg) * a - 20N + (3 kg)(-9.81 m/s^2)sin37 N = 0
a = -12.57 m/s^2
vf^2 = vi^2 + 2ad
0 = (5 m/s)^2 + 2 (-12.57 m/s^2) * d
d = 0.99 m
A 3-kg mass has a speed of 5.0m/s at the bottom of a 37 degree incline. How far up the incline does the mass slide before it stops if the friction force acting upon its motion is 20N?
Thanks.
1 answer