A force in the xdirection with magnitude Fx180N0530Nmx is ap

A force in the +x-direction with magnitude F(x)=18.0N(0.530N/m)x is applied to a 9.00 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.

Solution

Dtat is missing: take , initial position x = 0 and final position x = 17 m

work done = integral F(x). dx = 18*x - 0.53*x^2/2

apply the limits from x= 0 to 17, we get

    work done = 18*17 - 0.53*17^2/2

             = 229.415 J

from work-energy theorem, we have

work done = change in kinetic energy

   0.5*m*v^2 = 229.415

                  v = sqrt(2*229.415/9)

                     = 7.14 m/s