Question

a 3.00 kg object is moving in a plane, with its x and y coordinates given by x = 5t^2 - 1 and y = 3t^2 +2, where x and y are in meters and t is in seconds. find the magnitude of the net force acting on this object at t = 2.00 s

do i just substitute t?

Answer 1

is the answer \[3 \sqrt{136} ?\]

Answer 2

thats right

Answer 3

why are you asking me the answer o.O

Answer 4

i do not know how to do it and im asked the answer..that's bizarre lol

Answer 5

I thought u did have the answer! most questions give the final answer, that's why.

Answer 6

mehh not this one..

Answer 7

anyway, u don't need to substitute t at all. acceleration comes out to be a constant. multiply that with mass and u'll get the answer.

Answer 8

how do i find acceleration? 2d/t^2?

Answer 9

that's right. so u'll get the x and y components of accel.

Answer 10

hmm i cant visualize it...would you show a demo?

Answer 11

how abt u think of it as a projectile motion. we have constant acceleration. and displacement and velocity are also in 2D - x and y. so, instead of directly giving the values of disp and vel, the values of their components are given in terms of t. does this help?

Answer 12

ohhhhh definitely!!

Answer 13

so u got the sum?

Answer 14

so i will also have Fx and Fy?

Answer 15

yeah....u can go that way. and then Fnet = sqrt( Fx^2 + Fy^2)

Answer 16

oh of course

Answer 17

or else, when u ax and ay....calculate the net accel using the same formula.

Answer 18

wait...does t = 2 s matter? i differentiated x and i got a constant acceleration

Answer 19

*find

Answer 20

hm?

Answer 21

that's what i said b4. it doesn't matter.

Answer 22

yeah..just making sure

so Fx = 30 N
Fy = 18 N

so \[\sum F = \sqrt{8^2 + 30^2}?\]

Answer 23

yup. that's right. just that, under the square root, it should be 18^2...not 8^2

Answer 24

oh lol of course thanks :D

Answer 25

np :)