# 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 accidentally bumps the wheelbarrow. It slides...

Kinetic energy

Pete 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 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 of the ramp? (Remember the PE is a conserative force and the frictional force is nonconserative. The total KE is the difference between the two.)

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We need to find the height H of the top of the ramp. H = 50sin(20.0°) = 45.6 m.

The gravitational potential energy of the wheelbarrow at the top of the ramp is U = mgH. In our case, mg = 500 N, the weight of the wheelbarrow. So U = 500 N * 45.6 m = 22800 N*m. (Remember, that 1 N*m = 1 J).

If the wheelbarrow slid to the bottom without friction, it's kinetic energy would be equal to the starting potential energy, or 22.8 kJ. Instead, we need to take away the energy lost to friction. The work done by the friction force along the ramp path S is

W = ∫(F) dS = ∫(80.0 N) dS = 80S, evaluated from S = 0 to S = 50

W = 80.0 N * 50.0 m = 4000.0 J

So the total kinetic energy of the wheelbarrow at the bottom of the ramp is 22.8 kJ - 4.0 kJ = 18.8 kJ

The total force acting on the wheelbarrow is 500 N acting in vertically downward direction.

The component of this force along the ramp is causing the wheelbarrow to slide down. This force along the ramp

= 500*Sin20 = 500*0.3420 = 171 N

Part of this force is used for over coming the friction of 80 N.

The remaining force will cause the wheel brought o accelerate and thus gain kinetic energy. This force adding to kinetic energy is equal to

Force causing acceleration = 171 - 80 = 91 N

The total work done in accelerating the wheelbarrow

= (Force causing acceleration)*(Distance moved)

= 91*50 = 4550 J

The kinetic energy of wheelbarrow is same as the work done or energy used in accelerating the wheelbarrow.

Therefore kinetic energy of wheelbarrow at bottom of the ramp = 4550 J

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.