A rock of mass 32 kg accidentally breaks loose from the edge of a cliff and falls straight down. The magnitude of the air resistance that opposes its downward motion is 275 N. What is the magnitude of the acceleration of the rock?
(Weight - 275 N)/Mass = acceleration
Weight = 313.6 N
a = (313.6-275)/32 = 1.2 m/s^2
This accleration can only apply for an instant
Actually, acceleration will decrease and velocity and air resistance will increase until a limiting velocity is reached and acceleration becomes zero. This is a misleading problem.
To find the magnitude of the acceleration of the rock, we will first need to determine the net force acting on it. This can be done by subtracting the magnitude of the air resistance from the gravitational force.
The gravitational force acting on an object can be calculated using the formula:
Fg = m * g
Where:
Fg is the gravitational force,
m is the mass of the object, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the mass of the rock is given as 32 kg. So, the gravitational force can be calculated as:
Fg = 32 kg * 9.8 m/s^2
= 313.6 N
Now, since the object is falling straight down, the net force acting on it is the difference between the gravitational force and the air resistance. So,
Net force = Fg - air resistance
Net force = 313.6 N - 275 N
= 38.6 N
According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration:
Net force = mass * acceleration
Therefore, we can rearrange the equation to solve for the magnitude of the acceleration:
acceleration = net force / mass
acceleration = 38.6 N / 32 kg
= 1.20625 m/s^2
Therefore, the magnitude of the acceleration of the rock is approximately 1.21 m/s^2.