Solving Coefficient of Friction Error on Webassign

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The discussion revolves around a user encountering an error while calculating the coefficient of friction on Webassign. The user believes they have the correct force of friction and normal force values but is receiving a rejection from the program. It is identified that the normal force calculation may be incorrect, specifically suggesting the user used an incorrect value for gravitational acceleration. The correct gravitational acceleration should be 9.81 m/s² instead of 9.8 m/s². This adjustment could lead to the accurate calculation of the coefficient of friction.
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Hello,

I'm in the middle of a problem on the online program Webassign, trying to calculate the coefficient of friction in a problem. I have the correct value of the force of friction, and I believe I have the correct Normal Force value, but the program is rejecting my answer.

The problem involved a 21kg box sliding down a 38 degree inclined plane at an acceleration .270 m/s^2, and I calculated the force of friction to be 121.033N. To solve for the coefficient of friction I am using the equation Ff=uFn → u=Ff/Fn= 121.033N/205.8N=.588

Did I make an error or is it a glitch in the program?
 
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Your normal force is incorrect, it seems.
If you used N=mg, that's where your mistake lies.
 
ooh yikes, thank you, I didn't notice that. And I drew the vector diagram too, so I should've realized lol. Thank you!
 
Use g as 9.81m/s^2 and not 9.8m/s^2 and you might get the right answer
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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