Force, acceleration and mass

  • #1
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A constant force of 40 N acts on an object that is initially at rest on a smooth, horizontal surface. The object is observed to move a distance of 5.0m.

a.) what is the object's mass?
b.) what is the acceleration?

I can't seem to know what to do here. Can I solve this problem without given time. I am completely lost. can you please help me here
 

Answers and Replies

  • #2
Bystander
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Can I solve this problem without given time.
Can you write the general equations for mass and acceleration? Occasionally you will run into problems for which a specific numerical answer is not asked of you. You have stated that time is not given, so this might be such a problem.
 
  • #3
haruspex
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Are you quite sure there is no other information? Have you stated the problem word for word?
 
  • #4
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Yes, I have.. These are the exact words written in the book
 
  • #5
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Can you write the general equations for mass and acceleration? Occasionally you will run into problems for which a specific numerical answer is not asked of you. You have stated that time is not given, so this might be such a problem.
I don't know if this would really help but i'm dealing here "Newton's 2nd Law of Motion" so the formula here is F=ma..
 
  • #6
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Are you quite sure there is no other information? Have you stated the problem word for word?
Yes, I have. these are the exact words on the book
 
  • #7
jbriggs444
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Yes, I have. these are the exact words on the book
Then you should do as Bystander suggests. Treat time as an unknown. Label it t. Proceed to solve the problem. The result should be the object's mass expressed as function of t and its acceleration, expressed as a function of t.
 
  • #8
HallsofIvy
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You are not told how long it takes the object to move 5 meters? Then the best you can do is find the mass and acceleration as functions of that time.
 
  • #9
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would this mean..
vf=(5.0m/Time)
a=(5.0m / time^2)
m= F/a
m=40N time^2/5.0m

is this right??
 
  • #10
nasu
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Can you post a scan of the page in the book?
 
  • #11
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X(t)=0.5at^2 (the body is starting with no initial speed)=5(m)
a=10\t^2
F=40(N)=ma(Newton 2nd law)
by positioning of the value of acceleration we found in the new equation. and also transport members we found the mass
m=4t^2
the two values that you found dependent on the value of the time that it takes the body to move those 5 meters...
 
  • #12
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X(t)=0.5at^2 (the body is starting with no initial speed)=5(m)
a=10\t^2
F=40(N)=ma(Newton 2nd law)
by positioning of the value of acceleration we found in the new equation. and also transport members we found the mass
m=4t^2
the two values that you found dependent on the value of the time that it takes the body to move those 5 meters...
oh yeah.. i get it now.. i was a little confused with the formula a while ago.. THANKS A LOT FOR THE HELP ;)
 

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