Using SUVAT and Newton's Second Law

  • Thread starter Apothem
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Homework Statement



"A block of mass 10kg is pulled 20m up a smooth plane inclined at 45 degrees to the horizontal. The block is initially at rest and reaches a velocity of 2.0m/s at the tope of the plane. Calculate the magnitude of the force required, assuming it acts parallel to the plane."

Homework Equations



I am unsure about the question answer, but for my attempt I used v2=u2+2as, and rearranged it to find a, then used f=ma, to find the force.

The Attempt at a Solution



I drew a diagram with a horizontal line and a line at 45 degrees to that, I then used the equation as above, rearranged to find a as 0.1ms-2 (v=2 , u=0 , s=20, so I used the rearranged formula and got a=0.1ms-2). I then substituted this value for a into F=ma, with m being the mass of 10kg, however, the only thing I am unsure about is whether I have calculated the value of a correctly to begin with, then I can easily use F=ma. All help appreciated! EDIT: So the final force I calculated was 1N.
 
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Answers and Replies

  • #2
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Hi Apothem...

Welcome to PF!!!

Your value of acceleration is correct,assuming constant acceleration .But using F=Ma will give you F which is net force acting on the block,not the force which pulls the block up the inclined plane.

What are the forces acting on the block along the inclined plane ?

Edit:Are you required to calculate the net force or the force which pulls the block up the inclined plane ?
 
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  • #3
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Hi Apothem...

Welcome to PF!!!

Your value of acceleration is correct,assuming constant acceleration .But using F=Ma will give you F which is net force acting on the block,not the force which pulls the block up the inclined plane.

What are the forces acting on the block along the inclined plane ?

Thanks for the nice welcome Tanya!
That is all the question says word for word, which I put up in the first post. So....
EDIT: I think it is the force which pulls the block up the inclined plane.
 
  • #4
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Thanks for the nice welcome Tanya!
That is all the question says word for word, which I put up in the first post. So....
EDIT: I think it is the force which pulls the block up the inclined plane.

How did you get 0.05N ?
 
  • #5
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How did you get 0.05N ?

Sorry, I didn't mean 0.05, used a figure from a different question, so using F=ma (m=10kg , a =0.1ms-2, F=1N)
 
  • #6
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Sorry, I didn't mean 0.05, used a figure from a different question, so using F=ma (m=10kg , a =0.1ms-2, F=1N)

Okay..Now F=1N is the net force on the block .The net force is the resultant of all the forces acting on the block along the incline i.e parallel to the incline . But you need the force (assuming constant) that pulls the block .

So what are the forces acting on the block along the incline ?
 
  • #7
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Okay..Now F=1N is the net force on the block .The net force is the resultant of all the forces acting on the block along the incline i.e parallel to the incline . But you need the force (assuming constant) that pulls the block .

So what are the forces acting on the block along the incline ?

Sorry I think all I need to know is, the force parallel to the incline (because the question says: "Calculate the magnitude of the force required, assuming it acts parallel to the plane), because the question does not give you any forces acting on the block along the incline, so that means the only force we could calculate is 1N, so is this right? (for the force which acts parallel to the plane?).
 
  • #8
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What have you studied in the class about "Force" ? Have you touched upon FBD's (Free Body Diagrams) ?
 
  • #9
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What have you studied in the class about "Force" ? Have you touched upon FBD's (Free Body Diagrams) ?

Would it show all the forces acting upon the block?. So the weight would be vertically down the slope etc.
 
  • #10
1,540
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Weight(Force due to gravity) always acts vertically downwards ,not vertically down the slope.

Anyways...You didn't answer my question - Have you studied FBD's ?
 
  • #11
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Weight(Force due to gravity) always acts vertically downwards ,not vertically down the slope.

Anyways...You didn't answer my question - Have you studied FBD's ?

I meant the component of weight down the slope sorry, we did one question which used a FBD, however it was not explained.
 
  • #12
1,540
135
Well..in that case,it looks like the question is asking for net force . F=1N is the correct answer .
 
  • #13
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Well..in that case,it looks like the question is asking for net force . F=1N is the correct answer .

Thank you for all the help Tanya! I have given you a "Thanks" point, if that is even what they are called!
 
  • #14
1,540
135
Thank you for all the help Tanya! I have given you a "Thanks" point, if that is even what they are called!

You are welcome !

Thanks for the "Thanks" :smile:
 

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