(adsbygoogle = window.adsbygoogle || []).push({}); 7. A block of mass m = 5.5 kg is pulled up a θ = 22° incline as in the figure with a force of magnitude F = 38 N.

(a) Find the acceleration of the block if the incline is frictionless

So I got this one right, and here is what I did...

From the free body diagram I drew:

F_{net}= (38 N) + (-mg sin 22)

ma = 38 - mg sin 22

a = (38 - mg sin 22)/m

a = [ 38 - (5.5)(9.81)(sin 22) ] / (5.5)

a = (38 - 20.211) / 5.5

a = 3.24 m/s^{2}

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.

So...I know that

f_{k}= μ_{k}n

f_{k}= 0.12(F_{applied}sin θ - mg)

QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = F_{applied}sin θ + (- mg) after rotating my free body diagram so that I can see the axis better??? Because I almost got the answer...

f_{k}= 0.12[38 sin 22 - (5.5)(9.81)]

f_{k}= -4.767 (the negative number makes sense since kinetic friction is OPPOSING the motion of the block which is moving up the incline.

So then I add it to the equation for acceleration at the top

ma = (38 N) + (-mg sin 22) + (-4.767)

a = [ 38 - (5.5)(9.81)(sin 22) - 4.767 ] / (5.5)

a = 13.022 / 5.5

a = 2.36 m/s^{2}

This answer is wrong...Webassign says the answer is actuallya = 2.15 m/s^{2}

QUESTION 2: Where did I go wrong? Was it the normal force equation part?

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# Homework Help: Finding acceleration of a block on an incline without friction, then with friction!

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