Deep space object in zero gravity subject to forces

1. Jan 31, 2006

lomantak

The 1.4 kg object below is at rest, suspended in deep space, far from any planet or other body. If the forces shown below are exerted simultaneously on the object, what will be the velocity of the object after 3.5 s?

Split into x and y, so ...

For x:
$$[2.9cos(30)]-6.4$$
$$=-3.889$$

For y:
$$[2.9sin(30)]+4$$
$$=5.45$$

So take the sums of X and Y and do pythag, getting 6.6953 at -35.5 degrees as the resultant. Then taking the resultant force, plug it into $$F=ma$$? The question asks for velocity, but the formula $$F=ma$$ only has the acceleration. Any help is greatly appreciated! Thanks!

2. Jan 31, 2006

chroot

Staff Emeritus
You can find the acceleration, and you know the acceleration continues for 3.5 seconds. You should be able to find the final velocity by simply using:

$v(t) = v_0(t) + a t$

- Warren

3. Jan 31, 2006

Fermat

use v = u + at

where

v = final velocity
u = initial velocity
a = acceleration

4. Jan 31, 2006

cepheid

Staff Emeritus
1. Oooooh, careful! The y-component of the 2.9 N force is obviously acting downward (in the negative y-direction), yet you have not accounted for this. Keep in mind that the angle shown is, by convention, -30 degrees, or 330 degrees if you prefer. Using that value will give you the correct answer for both x-component and y-component.

2. Come on! Think! You have an initial velocity of zero. You have the acceleration, which is the rate at which the velocity changes with time. You also have a given time interval during which the object has that acceleration. Surely you know what the velocity will be at the end of that time interval...