A 6.6 kg body is at rest on a frictionless horizontal air track when a constant
ID: 1498925 • Letter: A
Question
A 6.6 kg body is at rest on a frictionless horizontal air track when a constant horizontal force vector F acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in the figure. The force vector F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force vector F between t = 0 and t = 1.5 s?Explanation / Answer
from equation of motion
(x-xo) = vo*t + (0.5*ax*t^2)
(0.8-0) = 0*t + (0.5*ax*2^2)
ax = 0.4 m/s^2
++++++
from work energy theorem
work done = change in KE
W = 0.5*m*(v^2-vo^2)
vo = 0
v = vo + ax*t = 0.4*1.5 = 0.6 m/s
W = 0.5*6.6*0.6^2
W = 1.188 J <<<-----answer