A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m/s² backward. (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 kg?

(c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.

• Force exerted by an opposing player = 800 N.
• The mass of the losing player plus equipment = 90.0 kg.
• Acceleration of the player =1.20 m/s².

Apply Newton’s second law of motion,

\({F_{{\rm{net}}}} = ma\)

\(F - f = ma\) ……………… (i)

Here,F_netis the net force, m is the mass of the artillery shell, f is the fiction force, F is the force exerted by opposing player, and a is the acceleration.

Substitute 90 kg for m, 800 N for F, and 1.20 m/s² for a in equation (i), and we get,

\(\begin{array}{c}800\;{\rm{N}} - f = 90\;{\rm{kg}} \times 1.2\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\800\;{\rm{N}} - f = 108\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\f = \left( {800 - 108} \right)\;{\rm{N}}\\f = 692\;{\rm{N}}\end{array}\)

Hence, the friction force between the losing player’s feet and grass is 692 N.

Substitute (110+90) kg for m, 692 N for f, and 1.20 m/s² for a in equation (i) and we get,

\(\begin{array}{c}F - {\rm{692}}\;{\rm{N}} = \left( {{\rm{110 + 90}}} \right)\;{\rm{kg}} \times {\rm{1}}{\rm{.2}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\F - {\rm{692}}\;{\rm{N}} = {\rm{240}}\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\F = \left( {{\rm{240 + 692}}} \right)\;{\rm{N}}\\F = {\rm{932}}\;{\rm{N}}\end{array}\)

Hence, the force exerted by the winning player is 932 N.

Draw the sketch of the system for part (a) as shown below:

Draw the sketch of the system for part (b) as shown below:

The net force equation is,

\({F_{{\rm{net}}}} = ma\).