Question

a student pushes a wall with a force of magnitude ( f_{s - w} ). the wall exerts a force of magnitude ( f_{w - s} ) on the student. the student then pushes harder on the wall, exerting a force of magnitude ( 2f_{s - w} ) on the wall. the magnitude of the force that the wall then exerts on the student is

a ( \frac{f_{w - s}}{2} )
b ( f_{w - s} )
c ( 2f_{w - s} )
d ( 4f_{w - s} )

Explanation:

Step1: Apply Newton's third law

Newton's third law states that for every action, there is an equal and opposite reaction. When the student pushes the wall with force \(F_{s - w}\), the wall pushes back with force \(F_{w - s}\), so \(F_{s - w}=F_{w - s}\).

Step2: Analyze the new force situation

When the student exerts a force \(2F_{s - w}\) on the wall, by Newton's third law, the force the wall exerts on the student \(F_{new}\) will satisfy \(F_{new}=2F_{s - w}\). But since \(F_{s - w}=F_{w - s}\), then \(F_{new} = 2F_{w - s}\).

Answer:

C. \(2F_{w - s}\)