41.3k views
13 votes
Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 64 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 69 kg and exerts an average force of 1367 N horizontally. (a) What is the acceleration (in m/s2 in the direction the heavy team is pulling) of the two teams

User DYZ
by
7.5k points

1 Answer

4 votes

Answer:


a=0.13m/s^2

Step-by-step explanation:

From the question we are told that

Mass of first team man
m_1=64kg

Force of man first team man
F_1=1350

Mass of second team man
m_2=69kg

Force of man second team man
F_2=1367N

Generally the equation for net force F_n is mathematically given by


F_n=9(m_1+m_2)a


9(m_1+m_2)a=9(f_2-f_1)


9(64+69)a=9(1367-1350)


a=(9(1367-1350))/(9(64+69))


a=0.127819m/s^2

Therefore the acceleration is given by


a=0.13m/s^2

User Trevortni
by
7.6k points