Calculating Acceleration of Block on Inclined Plane with Applied Force

[siderealtime]

A block slides down an inclined plane, here are the variables:

theta of incline = 37 degrees
mass of block = 10 kg
coefficient of kinetic friction = .500
applied force on block perpendicular to plane = 20N

\sum x = mgsin(theta) - fk = ma

\sum y = (20N + mgcos(theta)) - n = 0


n = 20N + mgcos(theta) = 98.3N
fk = \mu*n

I need to find acceleration of the block. Here is how I'm currently doing
it and I would like to know if this is correct.

solving for a in the first equation gives
a = mgsin(theta) - \muk*n / m

What I'm wondering is if the mass in the denominator is supposed to drop out?

 

[tiny-tim]

welcome to pf!

hi siderealtime! welcome to pf!

(have a mu: µ and a theta: θ and a sigma: ∑ and a degree: ° )

yes, that looks fine

(and the m won't drop out, because the 20N in this case isn't proportional to m)

 

[Xerxes1986]

Your normal force will certainly depend on mass, so it should cancel out.

 

[siderealtime]

Ugh, which is it, these both seem to make sense. Would anyone care to elaborate on how the division by m would go away or not?

 

[tiny-tim]

hi siderealtime!

(just got up :zzz: …)

xerxes is thinking of the usual normal force, which is proportional to m

but this 20N is an applied force, perhaps from something like a jet, whose value does not depend on m

(it would be different if say there was a 20N weight on top of the mass, which would contribute to the RHS of F = ma, but even then of course the weight would be vertical and not normal, so the proportions still wouldn't be preserved)

making m very large, for example, would make the 20N insignificant by comparison, and would considerably alter the acceleration

 

[siderealtime]

Thank you very much, Tim.