Question

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 , {kg} and exerts an average force of 1350 , {N} horizontally. Each of the second team’s members has an average mass of 73 , {kg} and exerts an average force of 1365 , {N} horizontally. (a) What is the magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?

The magnitude of the acceleration of the two teams is approximately:


a) 0.22 , {m/s}^2
b) 0.45 , {m/s}^2
c) 0.78 , {m/s}^2
d) 1.02 , {m/s}^2

Answer 1

The magnitude of the acceleration of the two teams is approximately 0.22 m/s². The tension in the section of rope between the teams is 6,075 N. The correct option is a. 0.22 , {m/s}^2

Step-by-step explanation:

To determine the magnitude of the acceleration of the two teams, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F=ma). We can calculate the total force exerted by both teams by adding up the individual forces of each team's members.

The total mass of the first team is 9 members x 68 kg = 612 kg. The total mass of the second team is 9 members x 73 kg = 657 kg. Thus, the total force exerted by both teams is 612 kg x 1350 N + 657 kg x 1365 N = 1,732,470 N.

To calculate the magnitude of the acceleration, we divide the total force by the total mass: a = F/m = 1,732,470 N / (612 kg + 657 kg) = 0.22 m/s². Therefore, the magnitude of the acceleration of the two teams is approximately 0.22 m/s²

To find the tension in the section of rope between the teams, we can use the equation T = F/2, where T is the tension and F is the force exerted by one team. We can calculate the force exerted by one team by multiplying the force per member by the number of team members: F = 1350 N x 9 = 12,150 N. Therefore, the tension in the section of rope between the teams is T = 12,150 N / 2 = 6,075 N.

The correct option is a. 0.22 , {m/s}^2