Asked by jdajaljk
A 10 kg block is allowed to slide down a ramp with uk = 0.15.
a. What is the value of the frictional force opposing the block's slide down the ramp?
b. What is the acceleration of the block?
a. What is the value of the frictional force opposing the block's slide down the ramp?
b. What is the acceleration of the block?
Answers
Answered by
drwls
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.
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.
Answered by
jdajaljk
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
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
Answered by
drwls
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.
It would be better to write it as 0.141 m/s^2. It is easier to read that way.
Answered by
Claire
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.
Answered by
John
I agree with Claire. My final answer for the acceleration was 1.97m/s^2
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