The figure gives the acceleration of a 4.0 kg particle as an applied force moves
ID: 1536739 • Letter: T
Question
The figure gives the acceleration of a 4.0 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 a_s = 10.0 m/s^2. How much work has the force done on the particle when the particle reaches x = 4.0 m, x = 7.0 m, and 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 x = 4.0 m, x = 7.0 m, and x = 9.0 m?Explanation / Answer
Given that, m = 4.0 kg
Now, work done = Force * distance
And the area under the graph would give acceleration !!
(a) From x = 0 to x = 4 ,
W = m * (1/2 * 10.0 * 1 + 10.0 * 3.0)
W = 4.0 * 35 J
W = 140.0 J
(b) When it reaches
W = m * (1/2 * 10.0 * 1 + 10.0 * 3.0 + 1/2*10.0*1 - 1/2*10.0*1 - 10.0*1)
W = 4.0 *(5 + 30 + 5 - 5 - 10)
W = 4.0*25 = 100.0 J
(c) W = m * (1/2 * 10.0 * 1 + 10.0 * 3.0 + 1/2*10.0*1 - 1/2*10.0*1 - 10.0*2 - 1/2*10.0*1 )
W = 4.0 * (5 +30 + 5 - 5 - 20 - 5)
W = 4.0 * 10 = 40 J
(d) We know,
Work done = Cahnge in potential Energy
put the values -
140 = 1/2*m*v^2
=> 140 = 1/2*4*v^2
=> v = 8.37 m/s
Like the above you can calculate (e) and (f).
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