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Friction & Acceleration

  • #1
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.
 

Answers and Replies

  • #2
Doc Al
Mentor
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m = 200 kg
f(app)= 700 N
U = 0.20
Ff = (0.20)(200)(-9.81)
=-392.4 N
Your error is giving g a minus sign. g is just the magnitude of the acceleration due to gravity.

So friction = μN = μmg. That's just the magnitude of the friction force. It's up to you to assign it the proper direction and sign; since it opposes the motion it will be negative (assuming the crate moves in the positive direction).

Set ΣF = ma and solve for a.

(In a nutshell: The friction reduces the effect of the applied force.)
 

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