Kinetic energyPete 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. Tammy...

The force of kinetic friction 80N when the  wheelbarrow is moving down acts on the the wheel barrow in an opposite direction of the movemement of the wheelbarrow.

 The component of the weIght force 500N  is 500N sin20 is acting along the inclination.

So the net force acting on the wheel barrorow = 500N sin20-80N.

Therefore the acceleration a of the wheel barrow is given by:

a = (500sin20-80)/(500/g).

So , using the equation of motion: v^2-u^2 = 2as, where v is final velocity and u is initial velocity , a is the acceleration and s is the dosplacement, and putting the values, u=0, v to be determined(if necessary), s = 50 and a = a = (500sin20-80)/(500/g), we get:

v^2-0 = 2{(500sin20-80N)/(500/g)}50 Or

(500/g)v^2 = 2*(500sin20-80)*50.

Dividing by 2, we get:

(1/2)500/g)v^2 = (500sin20-80)*50  = 4550.503583 Joules. Or

(1/2) mass*v^2 =  4550.503583 Joules.