What is the magnitude of the net force on a 3.40kg mass at t = 2.10 seconds?

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To determine the net force on a 3.40 kg mass at t = 2.10 seconds, first calculate the acceleration by finding the second derivatives of the position equations for x and y. The correct approach involves plugging in t = 2.10 to find ax and ay, then calculating the net acceleration as the vector sum of ax and ay, which are perpendicular. The final step is to multiply the resultant acceleration by the mass to find the net force in Newtons. Proper vector addition is crucial, as ax and ay cannot be simply summed directly.
Chuck 86
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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 don't know if I am supposed to plug in 2.10 for t at the beginning because i got a relativly different answer when i tried that
 
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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.
 
So i find the derrivative of the x and y then plug in t=2.10?
 
Yes. Find the derivative twice to get the acceleration.
 
i see thanks
 
After that i take the sum of the accelerations at 2.10 multiplied by the mass 3.40?
 
Chuck 86 said:
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.
 
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?
 
Chuck 86 said:
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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