A 6.5 kg particle starts from rest a x = o and moves under the influence of a single force
Fx = 6.1 + 7.6x - 1.1x^2, and Fx is n Newtons and x is in meters.
Find the power delivered to the particle when it is at x = 2.2m. answer in W.
Find the work done by this force on the particle as the particle moves from x = o m to x = 2.2m. answer in J.
Keep getting different answers. Would the work done equal the change in kinetic energy? Then how would I find the power delivered?
2006-10-04 07:20:52 · 2 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
Yes, work done would equal the change in kinetic energy, but it appears that the problem wants you to find it in a different way. I would integrate Fx from x=0 to x=2.2 to obtain work. (work = force * distance)
Power delivered to the particle can be calculated using
power = force * velocity
so, plug 2.2 into Fx to get the force, and using your answer for work (which should all now be kinetic energy), use the kinetic energy equation (KE = 1/2 * m * v^2) to obtain velocity.
good luck!
2006-10-06 21:20:41 · answer #1 · answered by buccinator 3 · 1⤊ 0⤋
Work done by force: integral of (6.1 + 7.6x - 1.1x^2 ) from x=0 to x= 2.2
F = 27.9N
v= (2K/m)^1/2= ((2*27.9N/6.5kg))^1/2=2.93m/s
d=v*t
t=d/v=2.2m/2.93m/s t=0.75s
P=F/t = 27.9N/0.75s = 37.16 watts
2014-03-03 02:09:15 · answer #2 · answered by Paul80 1 · 0⤊ 0⤋