A flat 100kg rock slides down a smooth rocky hill. It started from rest. The hill is 100m long and 50m high. The friction force between the rock and the ground is 150N. Use g=10m/s².
Calculate the speed of the rock at the bottom of the hill
1/2 m v^2 = m g h - friction work
50 v^2 = (100 * 10 * 50) - (100 * 150)
potential at top = m g h = 100 * 10 *50 = 50,000 Joules
work done by friction, force opposite to motion so negative = -150 *100
= -15,000 Joules
so
kinetic energy left at bottom = (1/2)m v^2 = 35,000 Joules
50 v^2 = 35,000
v^2 = 3500/5 = 700
To calculate the speed of the rock at the bottom of the hill, we can use the principle of conservation of energy. The total mechanical energy of the rock at the top of the hill (initial position) is equal to the total mechanical energy at the bottom of the hill (final position).
The total mechanical energy is given by the sum of the kinetic energy (KE) and the potential energy (PE).
At the top of the hill, all the potential energy is converted into kinetic energy, as the rock starts from rest. At the bottom of the hill, all the potential energy is converted back into kinetic energy.
The potential energy (PE) is given by the equation:
PE = m * g * h
where
m = mass of the rock (100 kg)
g = acceleration due to gravity (10 m/s²)
h = height of the hill (50 m)
PE = 100 kg * 10 m/s² * 50 m
PE = 50,000 J
At the bottom of the hill, all the potential energy (PE) is converted into kinetic energy (KE). So, the kinetic energy at the bottom is equal to the potential energy at the top.
KE = PE
KE = 50,000 J
The kinetic energy (KE) is given by the equation:
KE = (1/2) * m * v²
where
m = mass of the rock (100 kg)
v = velocity of the rock at the bottom
Plugging in the values, we have:
50,000 J = (1/2) * 100 kg * v²
Simplifying the equation:
50,000 J = 50 kg * v²
v² = 1,000 J / kg
v = √(1,000 J / kg)
v ≈ 31.62 m/s
Therefore, the speed of the rock at the bottom of the hill is approximately 31.62 m/s.