The figure gives the acceleration of a 4.0 kg particle as an applied force moves

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 (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) Number 105 Units (b) Number 75 Unites (c) Number 30 Units (e) Number 6.12 Units (f) Number 3.87 Units

Explanation / Answer

work done = Fs = ma s = m ( as)

therefore we need to find the area of curve,

a) W= 4 ( 0.5 ) ( 4+3) ( 3) (10)= 420 J

b) work done ( from x=0 x=7) =4x3x 10 = -120 j + 420 = 300 J

c) work done ( 0 to 9 ) = 4 x ( 3) (3)x 10 = -360 J + 300 = -60 J

d) work done = change in KE

  420 = 0.5 ( 4) v^2

v = 14.49 m/ s apprx

e) 420 - 120 = 0.5 ( 4) v^2

v = 12.247 m/s apprx

f) -60 J = 1/2 (4) v^2

v = -2.738 m/s apprx

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