Asmall box is held in place against a rough vertical wall bysomeone pushing on it with a force directed upward at 28 degreesabove horizontal. the coeffients of static and kineticfriction between the box and the wall are 0.40 and 0.30,respectively. the box slides down unless the applied forcehas a magnitude 23n. what is the mass of the box?

m = 1.9287 Kg

Explanation:

Given

∅ = 28º

μs = 0.40

μk = 0.30

F = 23 N

m = ?

We can apply  

∑Fh = 0  (+→)

N - Fh = 0    ⇒  N = Fh = F*Cos ∅

then    N = F*Cos ∅

∑Fv = 0  (+↑)

⇒    Fy - W + Ff = 0

where

Fy = F*Sin ∅

W = m*g

Ff = μs*N = μs*F*Cos ∅

then

⇒    Fy - W + Ff = F*Sin ∅ - m*g + μs*F*Cos ∅ = 0

⇒    m = (F / g)*(Sin ∅ + μs*Cos ∅)

⇒    m = (23 N / 9.81 m/s²)*(Sin 28º + 0.40*Cos 28º)

⇒    m = 1.9287 Kg

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.