Question
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.
Thanks.
Answers
plumpycat
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
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