An 15.0 −kg box is released on a 34.0 ∘ incline and accelerates down the incline at 0.240 m/s2?

(a)
Take g = 9.80 m s⁻²

Along the incline (Take all downward vectors to be positive) :
frictional force = -f
Component of gravitational force acting on the box= m g sin34.0°
Net force acting on the box, F = m g sin34.0° - f

F = m a
m g sin34.0° - f = m a
(15.0) (9.80) sin34.0° - f = (15.0) (0.240)
frictional force, f = (15.0) (9.80) sin34.0° - (15.0) (0.240) = 78.6 N


(b)
Component of gravitational force acting on the box = m g cos34.0°

Coefficient of kinetic friction
= f / (m g cos34.0°)
= 78.6 / [(15.0) (9.80) cos34.0°]
= 0.645