Question

A horizontal force of 12 newtons is applied to a 4.0 kg box that slides on a horizontal surface. The box starts from rest, moves a horizontal distance of 10 meters, and obtains a velocity of 5.0 m/s. The surface has friction. The friction force, assumed constant, is

Answer 1

From the conservation of energy:

$$Fd=F_fd+0.5mv^2$$

$$F_f=F-\frac{mv^2}{2d}$$

$$F_f=12-\frac{(4)(5)^2}{2(10)}=7\ N$$