A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the car has a speed of 25.6 m/s. Calculate the mechanical energy of the car at the top and bottom of the hill. (Assume the bottom of the hill has a height of 0 m, g=9.80 ms2/).

Explanation: Solution

1.

Gravitational potential energy

U=mgh=500*9.8*50

U=245000 J

2.

Kinetic energy is present at bottom of the hill

K=(1/2)mV2=(1/2)*500*27.82

K=193210 J

3.

Work done by friction

W=193210-245000=-51790 J

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.