The x and y coordinates of a 3.70-kg particle moving in the xy plane under the influence of a net force F are given by

Question

x = t4 − 5t
and
y = 7t2 + 1,
with x and y in meters and t in seconds. What is the magnitude of the force F at
t = 4.20 s?


N

Answer 1

forcex=t^4-5t=4.20^4-5(4.20)=290
forcey=7t^2+1=7*(4.90)^2 +1=169

force=sqrt(forcex^2+forcey^2)=336 N

Answer 2

Could some body show me how to get to the answer key 1290N? The x and y coordinates of a 4.30-kg particle moving in the xy plane under the influence of a net force F are given by
x = t4 − 5t
and
y = 5t2 + 1,
with x and y in meters and t in seconds. What is the magnitude of the force F at
t = 5.00 s?

Answer 3

You have to find the 2nd derivative of position (acceleration) and multiply that by the mass to find the x and y components, then take the sqrt of x^2+y^2 to find the total force.