When a 15.00 kg mass is attached to a vertical spring, the spring is stretched 2.0 m such that the mass is 6.0 m above the table
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a.
The gravitational force between two objects is given by:
where
G is the gravitational constant
M,m are the masses of the two objects
r is the separation between the objects
In this problem, we have:
M = 7000 kg is the mass of the elephant
m = 0.02 kg is the mass of the mouse
r = 1 m is the separation
Substituting into the equation, we find the gravitational force exerted by the elephant on the mouse:
b.
We can solve this part by keeping in mind Newton's third law of motion, which states that:
"When an object A exerts a force (action) on object B, then object B exerts and equal and opposite force (reaction) on object A".
In our problem, we can identify the elephant as object A and the mouse as object B: this means that the gravitational force exerted by the elephant on the mouse is the action, while the gravitational force exerted by the mouse on the elephant is the reaction. The two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the mouse on the elephant is equal to
where the negative sign simply means that the direction is opposite.