Consider a system of two masses and a pulley | vedclass

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

Question

(A) Let $m_1 = 0.8 	ext{ kg}$ and $m_2 = 0.2 	ext{ kg}$. The acceleration is $a = 5 	ext{ m/s}^2$. The coefficient of friction is $\mu = 0.1$.For t

Vedclass · Accepted Answer

$6.4$

Vedclass · Answer

(A) Let $m_1 = 0.8 	ext{ kg}$ and $m_2 = 0.2 	ext{ kg}$. The acceleration is $a = 5 	ext{ m/s}^2$. The coefficient of friction is $\mu = 0.1$.For the $0.2 	ext{ kg}$ block, the tension $T$ pulls it towards the pulley, and the friction force $f_2$ from the $0.8 	ext{ kg}$ block opposes this motion. The normal force on the $0.2 	ext{ kg}$ block is $N_2 = m_2 g = 0.2 	imes 10 = 2 	ext{ N}$.Equation of motion for $m_2$: $T - \mu N_2 = m_2 a \implies T - 0.1 	imes 2 = 0.2 	imes 5 \implies T - 0.2 = 1.0 \implies T = 1.2 	ext{ N}$.For the $0.8 	ext{ kg}$ block, the applied force $F$ acts to the right. The tension $T$ acts to the left. The friction force $f_1$ from the table acts to the left, and the friction force $f_2$ from the $0.2 	ext{ kg}$ block also acts to the left. The normal force from the table is $N_1 = (m_1 + m_2)g = (0.8 + 0.2) 	imes 10 = 10 	ext{ N}$.Equation of motion for $m_1$: $F - T - \mu N_1 - \mu N_2 = m_1 a \implies F - 1.2 - 0.1 	imes 10 - 0.1 	imes 2 = 0.8 	imes 5 \implies F - 1.2 - 1 - 0.2 = 4 \implies F - 2.4 = 4 \implies F = 6.4 	ext{ N}$.

Similar Questions

The coefficient of friction between blocks $A$ and $B$ is $\mu$. The minimum force $F$ with which $A$ must be pushed such that $B$ will not slip down is

Vedclass Test Series

Exam Paper Generator

Online Exam Module