Question

1. team a yanks the rope to the right with 135 n of force. team b refuses to lose and pulls left with 132.8 n. muddy ground friction adds 1 n resisting team as pull. the coach stands by, wondering if this is how ropes stretch. find the net force and direction.

Explanation:

Step1: Define positive - direction

Let the right - hand direction be positive. Team A's force $F_A = 135$ N, Team B's force $F_B=- 132.8$ N (negative because it is in the left direction), and the frictional force $F_f=-1$ N (resisting Team A's pull, so in the left direction).

Step2: Calculate the net force

The net force $F_{net}$ is the sum of all the forces acting on the rope. Using the formula $F_{net}=F_A + F_B+F_f$.
Substitute the values: $F_{net}=135+( - 132.8)+(-1)$.
$F_{net}=135 - 132.8-1$.
$F_{net}=1.2$ N.

Answer:

The net force is $1.2$ N to the right.