A big rock falls from a height towards water below, and hits water at a speed of 12 m/s. Rock was at a height of 15 m before falling and weighs 65 kg.
a) What was the average force of air resistance (e.g., friction) acting on the rock?
b) What is the force of friction underwater if the rock reaches a depth of 2.5 m before stopping and there is a buoyant force of 650 N [upward] acting on the rock once underwater.
a) To calculate the average force of air resistance acting on the rock, we can use the equation:
F = m * (v - u) / t
Where:
- F is the force of air resistance
- m is the mass of the rock (65 kg)
- v is the final velocity (12 m/s)
- u is the initial velocity (0 m/s)
- t is the time taken to reach the final velocity
First, let's find the time taken to reach the final velocity. We can use the equation of motion:
v = u + at
Since the rock is falling vertically, the initial velocity is 0 m/s and the acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the known values into the equation:
12 = 0 + 9.8t
Simplifying the equation:
9.8t = 12
t = 12 / 9.8
t ≈ 1.22 seconds
Now we can calculate the force of air resistance:
F = 65 * (12 - 0) / 1.22
F ≈ 634.43 Newtons
Therefore, the average force of air resistance acting on the rock is approximately 634.43 Newtons.
b) To find the force of friction underwater, we can use Newton's second law of motion:
F = ma
Where:
- F is the force of friction
- m is the mass of the rock (65 kg)
- a is the acceleration
In this case, the acceleration is determined by the difference between the buoyant force and the weight of the rock:
a = (Buoyant force - Weight of the rock) / m
a = (650 - mg) / m
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Weight of the rock = mg = 65 kg * 9.8 m/s^2
Now we can find the acceleration:
a = (650 - (65 * 9.8)) / 65
a ≈ 0.23 m/s^2
Finally, substitute the values into Newton's second law to find the force of friction:
F = 65 * 0.23
F ≈ 14.95 Newtons
Therefore, the force of friction underwater is approximately 14.95 Newtons.