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how do you set up the question?
I know that m= 25kg
First, you draw the free body diagram. And note that the question tells you: there is no acceleration(constant velocity) Then you apply Newton's second law to it.
ok so newtons second law you would use f=ma?
yes. \(\Sigma F = m\Sigma a\)
oh ok a
so we know that m=25 kg and what is the a?
first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25
well, since it is constant velocity, a=0.
for your a you should get 1.96 m/s^2
so how do you get 1.96 m/s^2
but the box is moving with constant velocity, i.e. no acceleration, right?
thats just a velocity acceleration is in m/s^2 not m/s i believe.
F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.
I hope this helps you , good luck and if by chance it's wrong I'm sorry!
Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.
so how do you find the force that is being exerted?
\(\Sigma F=ma\) Start with this picture. |dw:1352781084312:dw| As shown, the forces are F_x and F_f, of different directions. Applying the second law, \(F_x - F_f =ma=0\) as it is constant velocity.
Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.|dw:1352781249983:dw| Do you follow?
im still a bit confused about the Fn= Fg
Now, as \(F_g = m g\), where m=25kg and g=9.8, so you get \(F_g\) Now after you get \(F_g\) you can get \(F_N\) After you get \(F_N\), You get \(F_f\) Then since \( F_x-F_f=0\), \(F_x=F_f\).
On the y-axis, v=0,(it doesn't move vertically), so a=0 too. Since \(\Sigma F_y=F_N-F_g= 0\),it follows that \(F_N=F_g\)
oh ok i got you so far
Step by step, have you got the value of \(F_N\)?
im still confused on how to get Fn
i got Fg= 245
yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.
yeah, so next you will get the value of friction force...
oh ok so how do you find the friction force
\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.
nvm i mean (245)(.2)
yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.