Distance traveled by block on incline with friction

[Lori]

Was wondering if i got the answer correctly or got it by coincidence (answer is 3.0 m according to answer key)

1. Homework Statement
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Homework Equations

F = ma

The Attempt at a Solution

F_f = μ_k*N
N = mgcos(20)=18.41 = W

W = fd
W = ΔKE = 0-2 = -2
-2/f =d
f = just the x components since y component balance
f= F_{motion -}F_f
F_motion = mgsin20 = 6.7
F_f=0.4*N = 0.4*18.41=7.364
f = 6.7-7.364 = -0.664

d = -2/f = -2/-0.664 = 3.01 ≈ 3.0 m

So, the distance is 3.0 m

 

[stockzahn]

Coincidentally I've got the same result. Using the same method.

 

[haruspex]

f = 6.7-7.364

This line should raise a red flag. You are taking the difference of two numbers that only differ by about 10%, yet the precision of one of them is only given to two sig figs. If the range for that is really 6.65 to 6.75 then the error range in the result is ±10%. You should go back and ensure both inputs to this subtraction have at least three sig figs.

= mgcos(20)=18.41 = W
W = fd

Using W for two different things is most confusing.

Other than that, your method is sound.

 

[Lori]

This line should raise a red flag. You are taking the diffeence of two numbers that only differ by about 10%, yet the precision of one of them is only give to two sig figs. If the range for that is really 6.65 to 6.75 then the error range in the result is ±10%. You should go back and ensure both inputs to this subtraction have at least three sig figs.

Using W for two different things is most confusing.

Other than that, your method is sound.

Understood! I guess It's much more easier to be precise when I am writing down in pencil than typing on my computer! But, I see where the confusion is!

Thanks!