Question

Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-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 50.0 m. What is the wheelbarrow's kinetic energy at the bottom at of the ramp

Answer 1

Answer:

$E_{kinetic}=4551Joules$

Explanation:

The principe of work and energy is applied:

$E_{final}=E_{initial}+Work$

$E_{kinetic}=mgh-F*d$

The Work is negative because the Friction Force has a opposite direction as the displacement vector

The height is:

$h=d*sin\theta$

We replace this value in the Energy equation (mg=500N):

$E_{kinetic}=mg*d*sin\theta-F*d=500*50*sin20.0-80*50=4551Joules$