The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m.HELP?

Question

The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set by as = 9.0 m/s2. How much work has the force done on the particle when the particle reaches (a) x = 4.0 m, (b) x = 7.0 m, and (c) x = 9.0 m? What is the particle's speed and direction (give positive answer if the particle moves along x axis in positive direction and negative otherwise) of travel when it reaches (d) x = 4.0 m, (e) x = 7.0 m, and (f) x = 9.0 m?

NCS · Accepted Answer

work = area under force/distance curve, and we can convert the curve you are given into a f/d curve by multiplying the acceleration by the 3.0 kg mass.

So instead of scaling as = 9.0 m/s², we can scale Fs (at the same level) at 27.0 N.

The work done for 0 ≤ x ≤ 1 is ½*27*1 J = 14 J

and for 1 ≤ x ≤ 4 is 27*3 J = 81 J

so
(a) 0 ≤ x ≤ 4, the work is 81J+14J = 95 J

(b) For 4 ≤ x ≤ 6, the work is zero, and for 6 ≤ x ≤ 7, the work is -27 J, making the remaining total 68 J

(c) Subtract another (½+1)*27 = 41 J

leaving 27 J of work.

The net work is always positive for this graph, so the direction is always "positive." You can solve for the speed by rearranging
work = ½mv²

to yield
v = √(2*work / m)

(d) v = √(2*95/3) m/s = 8.0 m/s

(e) v = √(2*68/3) m/s = 6.7 m/s

(f) v = √(2*27/3) m/s = 4.2 m/s

Hope this helps!

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The figure gives the acceleration of a 3.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m.HELP?

1 Answer