Preston pushes a wheelbarrow weighing 490 N to the top of a 45.0-m ramp,inclined at 20.0° with the horizontal,and leaves it.Tamara accidentally bumps the wheelbarrow.It slides back down the ramp,during which an 80.0-N frictional force acts on it over the 45.0 m.What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²) A) 3,942 J B) 17,120 J C) 11,142 J D) 7,542 J E) 1,231 J

Question

Quizplus · Accepted Answer

The work done by Preston is:$$W_p=F_pd=490	ext{ N}(45.0	ext{ m})\cos180\degree=-22050	ext{ J}$$The work done by friction is:$$W_f=F_fd=-80.0	ext{ N}(45.0	ext{ m})=-3600	ext{ J}$$The change in gravitational potential energy is:$$U_g=mgh=490	ext{ N}(45.0	ext{ m})\sin20.0\degree=17120	ext{ J}$$The total work done on the wheelbarrow is:$$W=W_p+W_f+U_g=-22050	ext{ J}-3600	ext{ J}+17120	ext{ J}=-8330	ext{ J}$$The kinetic energy at the bottom of the ramp is:$$K=\frac{1}{2}mv^2=W=-8330	ext{ J}$$

Preston Pushes a Wheelbarrow Weighing 490 N to the Top