The figure shows a uniform rod of length 30 , | vedclass

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(D) The total mass of the rod is $M = 3.0 \,kg$ and total length is $L = 30 \,cm$.The linear mass density is $\lambda = \frac{M}{L} = \frac{3.0 \,kg}{

Vedclass · Accepted Answer

$24$

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(D) The total mass of the rod is $M = 3.0 \,kg$ and total length is $L = 30 \,cm$.The linear mass density is $\lambda = \frac{M}{L} = \frac{3.0 \,kg}{30 \,cm} = 0.1 \,kg/cm$.Mass of the $10 \,cm$ part is $m_1 = 0.1 	imes 10 = 1.0 \,kg$.Mass of the $20 \,cm$ part is $m_2 = 0.1 	imes 20 = 2.0 \,kg$.The net force on the system is $F_{	ext{net}} = 32 \,N - 20 \,N = 12 \,N$.The acceleration of the system is $a = \frac{F_{	ext{net}}}{M} = \frac{12 \,N}{3.0 \,kg} = 4 \,m/s^2$.Let $T$ be the force exerted by the $20 \,cm$ part on the $10 \,cm$ part. Considering the $10 \,cm$ part,the forces acting on it are $20 \,N$ (left) and $T$ (right).Applying Newton's second law: $T - 20 = m_1 a$.$T - 20 = 1.0 	imes 4$.$T = 20 + 4 = 24 \,N$.

The figure shows a uniform rod of length $30 \,cm$ having a mass of $3.0 \,kg$. The rod is pulled by constant forces of $20 \,N$ and $32 \,N$ as shown. Find the force exerted by the $20 \,cm$ part of the rod on the $10 \,cm$ part (all surfaces are smooth).

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