Question 1 The figure gives the acceleration of a 4.0 kg particle as an applied

Question

Question 1 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 as = 10.0 m/s. 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? as 2 x (m) -a (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number Units Units Units Units ( Units Units

Explanation / Answer

work done = m*area under graph

area of trapezium = (sum of parallel sides*distance betwen parallel side)/2

(a)

work done = m* area of trapezium = 4*(3+4)*10/2 = 140 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 = 4*[ (3+5)*10/2 - (2+1)*10/2) ] = 100 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 = 4*[ (3+5)*10/2 - (4+2)*10/2) ] = 40 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)*4*vf^2 = 140

vf = + 8.37 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)*4*vf^2 = 100

vf = + 7.07 m/s

---------------------

(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)*4*vf^2 = 40

vf = + 4.47 m/s

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