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

Answer:

$a=0.13m/s^2$

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

$F_n=9(m_1+m_2)a$

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

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

$a=\frac{9(1367-1350)}{9(64+69)}$

$a=0.127819m/s^2$

Therefore the acceleration is given by

$a=0.13m/s^2$