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 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

Answer 1

`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`