Question

d) hit go. describe the results of the tug of war match using physics terms (i.e. velocity, acceleration, displacement). refer to the sum of forces arrow (aka resultant force vector) in your answer.

Explanation:

Step1: Analyze forces

In a tug - of - war, two teams exert forces in opposite directions. The resultant force vector $\vec{F}_{net}=\sum\vec{F}$ is the difference between the forces exerted by the two teams. If team A exerts a force $\vec{F}_A$ and team B exerts a force $\vec{F}_B$, $\vec{F}_{net}=\vec{F}_A-\vec{F}_B$ (assuming $\vec{F}_A$ is in the positive direction).

Step2: Relate to acceleration

According to Newton's second law $\vec{F}_{net} = m\vec{a}$, where $m$ is the total mass of the rope and the people involved. If $\vec{F}_{net}
eq0$, there will be an acceleration $\vec{a}=\frac{\vec{F}_{net}}{m}$ in the direction of the resultant force.

Step3: Relate to velocity and displacement

If there is an acceleration $\vec{a}$ over a time interval $t$, the velocity $\vec{v}=\vec{v}_0+\vec{a}t$ (where $\vec{v}_0$ is the initial velocity, which is usually 0 at the start of the match). The displacement $\vec{d}=\vec{v}_0t+\frac{1}{2}\vec{a}t^{2}$. If $\vec{F}_{net} = 0$, the system is in equilibrium, $\vec{a}=0$, $\vec{v}$ remains constant (if it was 0 initially, it stays 0), and $\vec{d}=0$ (no movement). If $\vec{F}_{net}
eq0$, the team in the direction of the resultant force will start moving, with non - zero $\vec{v}$ and $\vec{d}$ in that direction.

Answer:

If the forces exerted by the two teams are equal, the resultant force $\vec{F}_{net}=0$, acceleration $\vec{a} = 0$, velocity $\vec{v}=0$ (assuming starting from rest), and displacement $\vec{d}=0$. If one team exerts a greater force than the other, the resultant force $\vec{F}_{net}$ is non - zero in the direction of the stronger team's force. This causes an acceleration $\vec{a}=\frac{\vec{F}_{net}}{m}$ in that direction. Over time, a non - zero velocity $\vec{v}=\vec{v}_0+\vec{a}t$ and displacement $\vec{d}=\vec{v}_0t+\frac{1}{2}\vec{a}t^{2}$ (with $\vec{v}_0 = 0$ initially) will develop in the direction of the stronger team's force.