In figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 n and the distance it acts over is 0.25 m, what is the final speed of the backpack?

I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!

The car would move to the right with 150 n of force. the 150 n of force from friction and air resistance slows the car from moving to the right with 300 n of force.