A horizontal force of F = 130 N is applied to a 35-kg cart across a 10-m level surface. If the cart accelerates at 2.3 m/s2, then what is the amount of work done by the force of friction as it acts to the motion of the cart?
work in =130 * 10 = 1300 Joules
a = 2.3
v = 2.3 t
d = 10 = (1/2) * 2.3 t^2
t^2= 20/2.3
t = 2.95 seconds
v = 2.3 * 2.95 = 6.78 m/s
Kinetic energy = (1/2) m v^2 = (1/2)(35)(6.78)^2
= 805 Joules
Energy lost to friction= 1300 - 805 = 495 Joules