The figure gives the acceleration of a 6.0 kg particle as an applied force moves
ID: 1642990 • Letter: T
Question
The figure gives the acceleration of a 6.0 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/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) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units (f) Number Units a(m/s2)Explanation / Answer
a) Average acceleration from x=0 to x=4 m
a4 = Area/length = ((0.5*1*10)+(3*10))/4 = 8.75 m/s2
Work done, W= force*distance = (m*a4)*x = 6*8.75*4 J = 210 J
b) Average acceleration from x=0 to x=7 m
a7 = Area/length = ((0.5*1*10)+(3*10)+(0.5*1*10)-(0.5*1*10)- (1*10))/7 = 3.6 m/s2
Work done, W= force*distance = (m*a7)*x = 6*3.6*4 J = 86.4 J
c) Average acceleration from x=0 to x=9 m
a9 = Area/length = ((0.5*1*10)+(3*10)+(0.5*1*10)-(0.5*1*10)- (2*10) - (0.5*1*10))/9 = 1.1 m/s2
Work done, W= force*distance = (m*a9)*x = 6*1.1*4 J = 26.4 J
d) velocity of the particle at x = 4 m
v = ( u2 + 2a4x)1/2 = (0+2*8.75*4)1/2 = +8.37 m/s
e) velocity of the particle at x = 7 m
v = ( u2 + 2a7x)1/2 = (0+2*3.6*4)1/2 = +5.4 m/s
f) velocity of the particle at x = 9 m
v = ( u2 + 2a9x)1/2 = (0+2*1.1*4)1/2 = +3.0 m/s