Use the law of conservation of energy to determine the final speed of the box if it starts from rest.?

Question

A gardener exerts a force of 1.5E2 N [22 derees below the horizontal] in pushing a large 18 kg box of flower seeds a distance of 1.6 m. The coefficient of kinetic friction between the box and the floor is 0.55.
The answer is supposed to be 1.4 m/s.

Answer 1

The normal force is g*18 + 150*sin(22deg) = 232.8 N, assuming g=9.81. Multiply this by friction coefficient 0.55 to find friction force = 128.0 N. The applied horizontal force is 150*cos(22 deg) = 139.1 N. The difference, 11.1 N, accelerates the box at 11.1/18 = 0.614 m/s^2. Use s=0.5*a*t^2 to find t = 2.28 s. V = a*t = 1.40 m/s.
Now using conservation of energy:
Applied energy = F hor * D = 139.1 * 1.6 m = 222.56 Nm. Friction energy = F fric * D = 128.0 * 1.6 m = 204.8 Nm. By conservation, kinetic energy KE = applied energy - friction energy = 17.76 Nm = 0.5 m*V^2. V=sqrt(2*17.76/18) = 1.40 m/s.