Acceleration of a 33 kg rock falling with 239 N air resistance

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shawonna23
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A rock of mass 33 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 239 N. What is the magnitude of the acceleration of the rock?

? m/s2
 
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This is really basic. Use F = ma:

F = ma

mg - 239 = ma

Now put in the values for m and g and solve for a.
 


The magnitude of the acceleration of the rock can be calculated using Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the gravitational force pulling the rock down and the air resistance pushing against it in the opposite direction. Therefore, the equation would be:

Fnet = ma

Where Fnet is the net force, m is the mass of the rock, and a is the acceleration. We know that the mass of the rock is 33 kg and the air resistance is 239 N, so we can rearrange the equation to solve for acceleration:

a = Fnet/m

Substituting the values, we get:

a = (mg - air resistance)/m

= (33 kg x 9.8 m/s2 - 239 N)/33 kg

= (323.4 N - 239 N)/33 kg

= 84.4 N/33 kg

= 2.56 m/s2

Therefore, the magnitude of the acceleration of the rock is 2.56 m/s2. This means that the rock is accelerating downward at a rate of 2.56 meters per second squared.