A rock of mass 40 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 271 N. What is the magnitude of the acceleration of the rock?

Question

anonymous · Answer

This problem is an example of Newton;s Second Law, F=ma

For free fall problems, the formula can be modified to be F= m*g*h

There is no height specified but the air resistance is given.

The air resistance is pushing up with a force of 271 N.

That is a vector in the upward direction and there must be vector of equal magnitude in the downward direction of 271 N.

With that information, Newtons Second Law can be used as follows:

F= 271 N

m= 40 kg

a=?

Therefore, 271 N = 40 kg * a

Keep in mind the units for a Newton is kg*m^2/s^2

so writing the units to see what cancels:

271 kg * m^2/s^2 = 40kg * a

271 kg * m^2/s^2  / 40 kg = a

The kilogram units cancel so you are left with M62/s^2 which is the appropriate unit for acceleration.

anonymous · Answer

...so what's the answer?