Question

How fast would a 15 kg box accelerate down a 25 degree snowy slope if the coefficient of kinetic friction were 0.15?

Answer 1

sum your forces in the direction PARALLEL with the incline:
= friction force - gravity of box
= u*mg*cos(25) - mg*sin(25)
= 0.15*15*9.81*cos(25) - 15*9.81*sin(25)
= -42.18

now, we know the box is moving, so that's why we never said "0=" when writting the equation above., so, we use a "-ma " negative becuase its moving down, ma becuase f=ma, and if we have the -42.18 as a unit of force, the other side of the equation must also be a unit of force., so we have:
-ma = -42.18
-15*a = -42.18
a = 42.18/15

boo ya