Magnitude of the net force

  • Thread starter Chuck 86
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  • #1
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A 3.40 kilogram mass is moving in a plane, with its x and y coordinates given by x = 4.85t2 - 1.05 and y = 3.25t3 + 2.10, where x and y are in meters and t is in seconds. Calculate the magnitude of the net force acting on this mass at t = 2.10 seconds.



I dont know if im supposed to plug in 2.10 for t at the begining because i got a relativly different answer when i tried that
 

Answers and Replies

  • #2
rl.bhat
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To find the force on the mass, first of all you have to find the acceleration of the mass.
Find ax = d^2x/dt^2 and ay = d^2y/dt^2 at given time t. Find net acceleration and then the net force.
 
  • #3
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So i find the derrivative of the x and y then plug in t=2.10?
 
  • #4
rl.bhat
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Yes. Find the derivative twice to get the acceleration.
 
  • #5
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i see thanks
 
  • #6
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After that i take the sum of the accelerations at 2.10 multiplied by the mass 3.40?
 
  • #7
rl.bhat
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After that i take the sum of the accelerations at 2.10 multiplied by the mass 3.40?
To find the net acceleration you have to find the vector sum of ax and ay.
 
  • #8
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i found that to be 9.7+ 40.95=50.65. do i take 50.65 multiplied by 3.40kg's to get the answer of the net force in newtons?
 
  • #9
rl.bhat
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i found that to be 9.7+ 40.95=50.65. do i take 50.65 multiplied by 3.40kg's to get the answer of the net force in newtons?
ax and ay are perpendicular to each other. You can't add then directly to get the resultant acceleration.
 

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