The figure gives the acceleration of a 3.0 kg particle as an applied force moves
ID: 1773742 • Letter: T
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 = 7.0 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?
a(m/s2)Explanation / Answer
work done = m*area under graph
area of trapezium = (sum of parallel sides)*distance betwen parallel sides/2
(a)
work done = m* area of trapezium = 3*(3+4)*7/2 = 73.5 J
(b)
work done =m*( area of trapezium from x = 0 to x = 5 - area of trapezium from x = 5 to x= 7)
Work done = 3*[ (3+5)*7/2 - (2+1)*7/2) ] = 52.5 J
(c)
work done =m*( area of trapezium from x = 0 to x = 5 - area of trapezium from x = 5 to x= 9)
Work done = 3*[ (3+5)*7/2 - (4+2)*7/2) ] = 21 J
(d)
from work energy theorem
work done = cahnge in KE
Wa = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)*m*vf^2 = Wa
(1/2)*3*vf^2 = 73.5
vf = + 7 m/s
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(e)
from work energy theorem
work done = cahnge in KE
Wb = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)*m*vf^2 = Wb
(1/2)*3*vf^2 = 52.5
vf = + 5.92 m/s
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(f)
from work energy theorem
work done = cahnge in KE
Wc = (1/2)*m*Vf^2 - (1/2)*m*vi^2)
vi = 0
(1/2)*m*vf^2 = Wc
(1/2)*3*vf^2 = 21
vf = + 3.74 m/s