Two blocks of masses m and 2m kept on a frict | vedclass

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

Question

(C) Let the acceleration of the system be $a$. The net force acting on the system is $F_{net} = F_2 - F_1 = 7.5t - 4.2t = 3.3t 	ext{ N}$.The total ma

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the acceleration of the system be $a$. The net force acting on the system is $F_{net} = F_2 - F_1 = 7.5t - 4.2t = 3.3t 	ext{ N}$.The total mass of the system is $M = m + 2m = 3m$.Using Newton's second law,$F_{net} = Ma$,we get $3.3t = (3m)a$,so $a = \frac{1.1t}{m}$.Now,consider the block of mass $2m$. The forces acting on it are $F_2$ in the forward direction and tension $T$ in the backward direction.Applying Newton's second law to this block: $F_2 - T = (2m)a$.Substituting the values: $7.5t - T = 2m \left( \frac{1.1t}{m} ight) = 2.2t$.Thus,$T = 7.5t - 2.2t = 5.3t$.Given that the tension $T = 10.6 	ext{ N}$,we have $5.3t = 10.6$.Solving for $t$,we get $t = \frac{10.6}{5.3} = 2 	ext{ s}$.

Similar Questions

Two blocks of $7\,kg$ and $5\,kg$ are connected by a heavy rope of mass $4\,kg$. An upward force of $200\,N$ is applied as shown in the diagram. The tension at the top of the heavy rope at point $P$ is ....... $N$ $(g = 10\,m/s^2)$.

$A$ monkey is descending from the branch of a tree with constant acceleration. If the breaking strength of the branch is $75 \%$ of the weight of the monkey,what is the minimum acceleration with which the monkey can slide down without breaking the branch?

What will be the acceleration of the system and tension in the rope connecting $5\,kg$ and $1\,kg$ blocks?

If the breaking strength of a rope is $\frac{4}{3}$ times the weight of a person,then the maximum acceleration with which the person can safely climb up the rope is ($g$ = acceleration due to gravity).

Vedclass Test Series

Exam Paper Generator

Online Exam Module