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Natasha_Volkova [10]
3 years ago
5

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
Physics
1 answer:
Serggg [28]3 years ago
4 0

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

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