Yes, 86.2 N sounds like a more accurate answer. Great job on your solution!

  • Thread starter Medgirl314
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  • #1
Medgirl314
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Homework Statement



A 15 kg crate rests on the floor. The coefficient of friction between the crate and the floor is 0.28. How much horizontal force is needed to accelerate the crate at 3.0 m/s^2?


Homework Equations



Fnet=ma

The Attempt at a Solution


m=15 kg
F-f=ma
F-[itex]M[/itex]N=ma
F-[itex]M[/itex]mg=ma
F=ma+[itex]M[/itex]mg
F=15*3+.38*15*9.8
F=81.16 N.
For a final rounded answer of: F=81.2 N.

This answer seems correct to me, could someone please confirm?

Thanks! :smile:
 
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  • #2
Medgirl314 said:

Homework Statement



A 15 kg crate rests on the floor. The coefficient of friction between the crate and the floor is 0.28. How much horizontal force is needed to accelerate the crate at 3.0 m/s^2?


Homework Equations



Fnet=ma

The Attempt at a Solution


m=15 kg
F-f=ma
F-[itex]M[/itex]N=ma
F-[itex]M[/itex]mg=ma
F=ma+[itex]M[/itex]mg
F=15*3+.38*15*9.8 (0.28 is the μ in the prob. statement)
F=81.16 N.
For a final rounded answer of: F=81.2 N.

This answer seems correct to me, could someone please confirm?

Thanks! :smile:

I don't get your result, even with μ = 0.28. Try your calculator again.
 
  • #3
Oops, I thought I corrected that, sorry! Weird, I somehow mixed up the digits. Does 86.16, rounded to 86.2, sound better?

Thanks!
 
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