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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

Sagot :

okpalawalter8

Answer:

[tex]a=0.13m/s^2[/tex]

Explanation:

From the question we are told that

Mass of first team man [tex]m_1=64kg[/tex]

Force of man first team man [tex]F_1=1350[/tex]

Mass of second team man [tex]m_2=69kg[/tex]

Force of man second team man [tex]F_2=1367N[/tex]

Generally the equation for net force F_n is mathematically given by

[tex]F_n=9(m_1+m_2)a[/tex]

[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]

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

[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]

[tex]a=0.127819m/s^2[/tex]

Therefore the acceleration is given by

[tex]a=0.13m/s^2[/tex]

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