# work

• Oct 3rd 2008, 11:58 PM
winterwyrm
work
The only force acting on a 3.8 kg canister that is moving in an xy plane has a magnitude of 4.7 N. The canister initially has a velocity of 4.2 m/s in the positive x direction, and some time later has a velocity of 4.0 m/s in the positive y direction. How much work is done on the canister by the 4.7 N force during this time?

Note: I'm sure it has something to do with the work-kinetic energy theorum. thx
• Oct 4th 2008, 01:54 AM
Greengoblin
Work is given by $W=Fs$ where F = force and s = the distance over which the force is applied (for a changing force it's an integral).

So you need to find the distance. First find the acceleration using F=ma.
$
a=\frac{F}{m} = \frac{4.7}{3.8} =$

Then using equations of motion, $v^2=u^2+2as$ you can get the distance and then the work. (v is final velocity, u is initialvelocity, a is acceleratio, and s is the distance)

You should be able to do the calculations. :)
• Oct 4th 2008, 09:09 AM
skeeter
since F = 4.7 N is the only force acting on the object, it does all all the work which happens to also be the net work on the object.

so, yes, it has everything to do with the work-KE theorem.

$W_{net} = \Delta KE = \frac{1}{2}m(v_f^2 - v_0^2)$
• Oct 4th 2008, 10:00 AM
winterwyrm
... Oh no, these two methods give two different results... Or maybe I'm just messing something up...