Asked by Lena
A 200kg crate us pushed horizontally with a force of 700 N. If the coefficient of friction is 0.20, calculate the acceleratio of the crate.
I tried doing this but I don't seem to be getting the right answer.
m = 200 kg
f(app)= 700 N
U = 0.20
Ff = (0.20)(200)(-9.81)
=-392.4 N
Fnet = 700-(-392.4)
= 1092.4 N
Fnet = mass x acceleration
1092.4 = 200 (a)
5.462 m/s^2 = a
The answer booklet says that I should be getting 3.3m/s^2 as my answer.
I tried doing this but I don't seem to be getting the right answer.
m = 200 kg
f(app)= 700 N
U = 0.20
Ff = (0.20)(200)(-9.81)
=-392.4 N
Fnet = 700-(-392.4)
= 1092.4 N
Fnet = mass x acceleration
1092.4 = 200 (a)
5.462 m/s^2 = a
The answer booklet says that I should be getting 3.3m/s^2 as my answer.
Answers
Answered by
bobpursley
You should add forces to get net force
netforce=fapplied+friction and then fricion force is a negative, so
netforce=700-392.4
netforce=fapplied+friction and then fricion force is a negative, so
netforce=700-392.4
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