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 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?
5 answers
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
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.
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.
I agree with Claire. My final answer for the acceleration was 1.97m/s^2
1 answer