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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$24$

Vedclass · Answer

(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).

Similar Questions

$A$ uniform rope of mass $1.0 \ kg$ is connected to a box of mass $2.0 \ kg$,which is placed on a smooth horizontal surface. The free end of the rope is pulled horizontally by a force of $6 \ N$. Find the tension at the midpoint of the rope. ............ $N$

$A$ force $F = 10t$ acts on a block of mass $m = 10 \, kg$ as shown in the figure. The coefficient of friction is $\mu = 0.4$. Find the time $t$ when the block loses contact with the surface.

$A$ rope of mass $5 \,kg$ is moving vertically in a vertical position with an upward force of $100 \,N$ acting at the upper end and a downward force of $70 \,N$ acting at the lower end. The tension at the midpoint of the rope is ......... $N$.

Two blocks are in contact on a frictionless table. One has mass $m$ and the other $2\,m$. $A$ force $F$ is applied on $2\,m$ as shown in the figure. Now,the same force $F$ is applied from the right on $m$. In the two cases respectively,what is the ratio of the contact force between the two blocks?

With what minimum acceleration can a fireman slide down a rope if the breaking strength of the rope is $\frac{2}{3}$ of his weight?

Vedclass Test Series

Exam Paper Generator

Online Exam Module