The figure gives the acceleration of a 5.0 kg particle as an applied force moves
ID: 1533967 • Letter: T
Question
The figure gives the acceleration of a 5.0 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 (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, (f) x = 9.0 m?Explanation / Answer
a) Work done = area under the graph * mass
For 0-4s
Area = 0.5(3+4)*10 = 35 units
W = 35*5 = 175 J
b) For 0-7 s
We can divide it into two parts 0-5s and 5-7s
Area 0-5s = 0.5*10*(3+5) =40 units
Area 5-7s = 0.5*10*(1+2) = 15 units (negative)
Total area = 25 units
Total work done = 25*5 = 125 J
c) For 0-9
We can divide it into 0-5 s and 5-9s
Area 0-5 = 40units
5-9 = 0.5*10(2+4) =30 units
Net area = 40-30 = 10units
Work done = 10*5 = 50J
Now to calculate velocity
Work done = change in KE = 0.5m(v2 - u2)
Since u = 0 so KE = 0.5mv2
d) For 0-4s
175 = 0.5*5v2
v = 8.367 m/s
e) for 0-7s
125 = 0.5*5v2
v = 7.07 m/s
f) for 0-9s
40 = 0.5*5v2
v = 4 m/s