Question

A person standing on a bridge at a height of 115 m above a river drops a 0.250-kg rock. (a) What is the rock’s mechanical energy at the time of release relative to the river surface? (b) What are the rock’s kinetic, potential, and mechanical energies after it has fallen 75.0 m? (c) Just before the rock hits the water, what are its speed and total mechanical energy?

Answer 1

(a) The energy relative to the river surface is a potential energy. It can be calculated as $E_p = mgh,$ where g is gravitational acceleration and h is the height above the river surface. Therefore,

$E_{p,1} = 0.250\,\mathrm{kg}\cdot9.81\,\mathrm{N/kg}\cdot115\,\mathrm{m} \approx 282\,\mathrm{J}.$

We should note that in the initial moment the kinetic energy is equal to 0, so the total energy is $E=E_{k,1}+E_{p,1} = 0+E_{p,1} = E_{p,1} = 282\,\mathrm{J}.$

(b) After the rock has fallen 75.0 m its height above the river surface is $115-75 = 40\,\mathrm{m}.$ Therefore, the potential energy relative to the surface of river is

$E_{p,2} = mgh_2 = 0.250\,\mathrm{kg}\cdot9.81\,\mathrm{N/kg}\cdot40\,\mathrm{m} \approx 98\,\mathrm{J}.$

According to the conservation of energy law, the total energy is still 282 J, therefore, the kinetic energy is $E_{k,2}=E-E_{p,2} = 282\,\mathrm{J} - 98\,\mathrm{J} = 184\,\mathrm{J}.$

(c) Just before the hit the potential energy will turn to 0 and kinetic energy ill be equal to the total energy. Therefore, $E_{p,3} = 0, \;\; E_k = E - E_{p,3} = E - 0 = 282\,\mathrm{J}.$

The kinetic energy is $E_{k,3} = \dfrac{mV_3^2}{2}, \;$ therefore, the speed is $V_3 = \sqrt{\dfrac{2E_{k,3}}{m}} = \sqrt{\dfrac{2\cdot282\,\mathrm{J}}{0.250\,\mathrm{kg}}} = 47.5\,\mathrm{m/s}.$