91Ó°ÊÓ

• 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,

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