Another seemingly easy forces question

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A block is pushed across a surface with a coefficient of kinetic friction of 0.15 using a 150N force, resulting in an acceleration of 2.53m/s². The discussion revolves around calculating the block's mass, with participants struggling to apply the correct equations due to confusion over force direction and significant figures. The correct approach involves recognizing that friction opposes the applied force, leading to the equation ma = Fa - Ff. After several attempts, the correct mass is determined to be approximately 37.5kg, with a consensus on rounding to 38kg for simplicity. The importance of significant figures in reporting the final answer is also emphasized.
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Posting this from my phone, apologies.
"A block is pushed across a horizontal surface with a coefficient of kinetic friction of 0.15 by applying a 150N horizontal force. If the block accelerates at 2.53m/s^2 find the mass of the block."

I don't even know where to start. I know the force applied, acceleration and kinetic friction coefficient. I can't use Fnet=ma or Fnet=sum of all forces because I don't have the mass or Fnet. I can't calculate friction without mass (since normal force must be equal to gravity in this scenario, and is there equal to mg). I've drawn a fBD but that didn't really help. Can I get a hint on where to start? Perhaps a system of equations?
 
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Element1674 said:
Posting this from my phone, apologies.
"A block is pushed across a horizontal surface with a coefficient of kinetic friction of 0.15 by applying a 150N horizontal force. If the block accelerates at 2.53m/s^2 find the mass of the block."

I don't even know where to start. I know the force applied, acceleration and kinetic friction coefficient. I can't use Fnet=ma or Fnet=sum of all forces because I don't have the mass or Fnet. I can't calculate friction without mass (since normal force must be equal to gravity in this scenario, and is there equal to mg). I've drawn a fBD but that didn't really help. Can I get a hint on where to start? Perhaps a system of equations?

Yes. A system of equations. Write the equations down in terms of the unknown mass 'm' and try to solve for it.
 
Yeah I can't quite seem to get it even knowing that. I tried using:
Fnet=ma and Fnet= Ff+Fa
Since they both equal Fnet, they must be equal:
Ma=Ff+Fa
Ma=(mu)mg+Fa
Then I get stuck. I tried rearranging and factoring out mass and I get 58kg (the answer is 38kg)
 
Element1674 said:
Yeah I can't quite seem to get it even knowing that. I tried using:
Fnet=ma and Fnet= Ff+Fa
Since they both equal Fnet, they must be equal:
Ma=Ff+Fa
Ma=(mu)mg+Fa
Then I get stuck. I tried rearranging and factoring out mass and I get 58kg (the answer is 38kg)

You need to keep track of the direction of the forces. Do the frictional force and the applied force both point in the same direction as the acceleration?
 
Element1674 said:
Yeah I can't quite seem to get it even knowing that. I tried using:
Fnet=ma and Fnet= Ff+Fa
Since they both equal Fnet, they must be equal:
Ma=Ff+Fa
Ma=(mu)mg+Fa
Then I get stuck. I tried rearranging and factoring out mass and I get 58kg (the answer is 38kg)
Maybe you're confusing yourself by using two symbols for the same mass. All you have to do is solve your equation for m. Also be careful of the signs of the forces.
 
Well acceleration and the force applied are in the same direction. Friction is opposite to that. I don't understand how that changes my calculations though :/
 
Element1674 said:
Well acceleration and the force applied are in the same direction. Friction is opposite to that. I don't understand how that changes my calculations though :/

It would change ma=mu*m*g+Fa to ma=Fa-mu*m*g. That's a difference.
 
Ok yes I did that and I got 56kg as an answer. Still incorrect :/
 
Element1674 said:
Ok yes I did that and I got 56kg as an answer. Still incorrect :/

Put the numbers in and show how you got it. That's not what I get.
 
  • #10
Wait are you getting 37.485...kg? I tried agAin and that's what I got
 
  • #11
Element1674 said:
Wait are you getting 37.485...kg? I tried agAin and that's what I got

Yes, that's about right.
 
  • #12
Element1674 said:
Wait are you getting 37.485...kg? I tried agAin and that's what I got
Yes, but you have too many significant figures.
 
  • #13
Yay I got it! :) thank you so much!
 
  • #14
With significant figures it is 37 right? I think the textbook just rounded wrong
 
  • #15
Element1674 said:
With significant figures it is 37 right? I think the textbook just rounded wrong

Depends exactly what you use for g. If you use 9.8m/s^2 you get 37.5kg. Round up, round down? Not really important.
 
  • #16
Element1674 said:
With significant figures it is 37 right? I think the textbook just rounded wrong
You could probably get away with 37.5, since the acceleration has 3 figures, the the force has 3, too, arguably.
 
  • #17
The blocks acceleration is 2.53m/s^2 which is three significant figures so an answer of 37.5kg is correct. Just don't forget the correct units.
 
  • #18
Oh ok. I've learned significant figures differently. 0.15 (the coefficient of friction, can't scroll up and see but i think that's the value) has 2 significant figures and so does the force (150N). The acceleration has 3 but you go by the least significant term so it has 2 in the end
 
  • #19
Element1674 said:
Oh ok. I've learned significant figures differently. 0.15 (the coefficient of friction, can't scroll up and see but i think that's the value) has 2 significant figures and so does the force (150N). The acceleration has 3 but you go by the least significant term so it has 2 in the end

You said the book gave 38kg. So they must be rounding 37.5kg up to 38kg. This is really all much less important than getting the physics right.
 
  • #20
The force thought with 150N has three significant figures. Leading zeros are not included in significant figures but trailing zeros are. The right answer was found but on a test it is always best to account for significant figures and I have always been expected too. Here is a link on a brief overview of significant figures.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html
 
  • #21
I am getting so confused. I will leave it as 2 signficant figures. Thanks all
 
  • #22
Element1674 said:
Oh ok. I've learned significant figures differently. 0.15 (the coefficient of friction, can't scroll up and see but i think that's the value) has 2 significant figures and so does the force (150N). The acceleration has 3 but you go by the least significant term so it has 2 in the end
You are correct. I said you could "get away with" using 3 figures because there is a little bit of wiggle room, but strictly you should only use 2 here.
 
  • #23
SpaceDreamer said:
The force thought with 150N has three significant figures. Leading zeros are not included in significant figures but trailing zeros are.
That is not exactly correct; trailing zeros on integers are ambiguous. That is why scientific notation is preferred; trailing zeros after the decimal point are significant.
 

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