Posted by jdajaljk on Friday, January 28, 2011 at 6:04pm.
After you have learned how to compute the frictional force, and applied
F(net) = ma in the downward direction along the ramp, feel free to post your work for critical evaluation.
We are not going to do it for you.
a.Fk=(u)Fn
Fk=(0.15)9.40=1.41 N
10cos(20)=9.40=Fn
b.Fnet=(mass)a
1.41 N=(10 kg) a
a=.141 m/s^2
Good work. I almost didn't see the decimal point in front of .141
It would be better to write it as 0.141 m/s^2. It is easier to read that way.
I don't understand why the normal force is 9.4. If the object weighs 10 kg, the Fg in the y direction should be 92.1N, correct which would make the Fn 92.1N as well.
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