posted by Terrence Eko .
A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s^2.
Find coefficient between the box and the ramp.
What acceleration would a 175 kg box have on this ramp?
The net force down the ramp is
(mass) x (acceleration) = 75*3.60 = 270 N
The component of weight down the ramp is
M g sin 25 = 310.6 M
The friction force is the difference between those two numbers, 40.6 N
The kinetic friction coefficient is
(friction force)/(normal weight component)
= 40.6/[735 cos 25] = 0.06
a)F=ma Fg=mg 310.9-270 cos(25)735
75(3.60) 75(9.81) 41N 667N
F parallel= sin(25)735=310.9N