Question

A person kicks a 4.0-kilogram door with a using the door to accelerate at 12 meters per seconds. What is the magnitude of the force exerted by the door on the person?

Answer 1

`|vecF|=48N`

Explanation:

A commonly used statement of Newton's third law: every action (force) produces and equal and opposite reaction. Therefore, the magnitude of the force exerted by the person on the door is equal to the magnitude of the force exerted by the door on the person.

We can calculate the force the person exerts on the door using Newton's second law:

`vecF=mveca`

`vecF=(4.0kg)(12m/s)`

`vecF=48(kgm)/s^2=48N`

The force exerted on the door by the person is `48N`, which means that force exerted on the person by the door is `-48N`, or `48N` in the opposite direction.

Therefore, the magnitude of the force exerted on the person by the door is `48N`.