What is The New velocity of the box?

  • Thread starter ohheytai
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  • #1
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Homework Statement


A 24 kg box is being pushed across the floor by a constant force ‹ 107, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.18. At t = 8.0 s the box is at location ‹ 11, 2, −4 › m, traveling with velocity ‹ 5, 0, 0 › m/s. What is its position and velocity at t = 9.7 s?



Homework Equations





The Attempt at a Solution


i got <12.11,0,0> and <7.58,0,0> but they were both wrong
 

Answers and Replies

  • #2
rock.freak667
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Since the force is in the x-direction, you need to get the frictional force which would be Frictional force = <μmg,0,0>, then find the resultant force,F. (and F=ma :wink: )
 
  • #3
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yea but i still dont get how to find v final at 9.7s
 
  • #4
rock.freak667
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yea but i still dont get how to find v final at 9.7s

Using my suggestion, once you get the resultant force vector and apply F=ma, you can get the vector a.

Then you can use v(t) = ∫a(t) dt
 
  • #5
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can you show me how to get the answer?
 
  • #6
rock.freak667
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can you show me how to get the answer?

I can't pose the solution for you, but I will start you off.

Resultant force, ma(t) = Constant force - frictional force
 

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