Answer to Question #236789 in Mechanics | Relativity for Anas

Question #236789

A 9.00-kg hanging object is connected by a light,inextensible cord over a light ,frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as -.200, find the tension in the string


1
Expert's answer
2021-09-13T15:44:06-0400


the arrow in brown shows the friction,

block m2:

m2gT=m2a9×9.8T=9a      (1)Nm2g=0N=m2g=5×9.8=49  Nf=μNf=0.2×49=9.8  Nm_2 g - T = m_2 a \\ 9 \times9.8 - T = 9a \;\;\; (1) \\ N - m_2 g = 0 \\ N = m2 g = 5 \times 9.8 = 49 \; N \\ f = μN \\ f = 0.2 \times 49 = 9.8 \;N

block m1:

Tf=m1aT9.8=5a      (2)T - f = m_1a \\ T - 9.8 = 5a \;\;\;(2)

solving 1 and 2:

T=37.8  Na=5.6  m/s2T = 37.8 \; N \\ a = 5.6 \;m/s^2


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