A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating backward at 1.20 m/s².
(a) What is the force of friction between the losing player's feet and the grass?
(b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110.0 kg?

According to newton's second law of motion, acceleration of an object is directly proportional to the net force acting on it. When there is no net force force acting on the body, there is no acceleration. A force is a push or a pull, and the net force ΣF is the total force, or sum of the forces exerted on an object in all directions.

[tex]F_{net}[/tex] ∝ a

[tex]F_{net}[/tex] = ma

[tex]F_{applied} - F_{fric}[/tex] = ma

Given data:

[tex]F_{applied}[/tex] = 800 N

Mass = m = 90 kg

acceleration = a = 1.2 m/s²

[tex]F_{fric}[/tex] = ?

800 - [tex]F_{fric}[/tex] = (90)(1.2)

[tex]F_{fric}[/tex] = 692 N

b)

According to newton's second law of motion,

[tex]F_{net}[/tex] ∝ a

[tex]F_{net}[/tex] = ma

[tex]F_{applied} - F_{fric}[/tex] = ma

Given data:

If we assume the same friction and acceleration between player's feet and ground as calculated in part a

[tex]F_{fric}[/tex] = 692 N

acceleration = a = 1.2 m/s²

We take the equal mass to the total mass of both the players because when the winning player push losing player backward, he exert force on the ground not only due to his mass but also due to the mass of losing player.

Mass = M = m₁ + m₂ = 110 kg + 90 kg

= 200 kg

[tex]F_{applied}[/tex] = ?

[tex]F_{applied}[/tex] - 692 N = (200)(1.2)

[tex]F_{applied}[/tex] = 692 + 240

[tex]F_{applied}[/tex] = 932 N