The coefficient of friction between 4 , kg an | vedclass

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

Question

(C) Let $m_1 = 4 \, kg$ (top block $P$) and $m_2 = 5 \, kg$ (bottom block $Q$).Coefficient of friction between $P$ and $Q$ is $\mu_1 = 0.2$. Coefficie

Vedclass · Accepted Answer

Maximum acceleration of $4 \, kg$ block is $2 \, m/s^2$.

Vedclass · Answer

(C) Let $m_1 = 4 \, kg$ (top block $P$) and $m_2 = 5 \, kg$ (bottom block $Q$).Coefficient of friction between $P$ and $Q$ is $\mu_1 = 0.2$. Coefficient of friction between $Q$ and ground is $\mu_2 = 0.1$.Maximum static friction between $P$ and $Q$ is $f_{1,max} = \mu_1 m_1 g = 0.2 	imes 4 	imes 10 = 8 \, N$.Maximum static friction between $Q$ and ground is $f_{2,max} = \mu_2 (m_1 + m_2) g = 0.1 	imes (4 + 5) 	imes 10 = 9 \, N$.For the system to move,the applied force $F$ must exceed $f_{2,max} = 9 \, N$. Thus,minimum force is $9 \, N$. Statement $A$ is incorrect.If $F = 4 \, N Maximum acceleration of $P$ is $a_{max} = f_{1,max} / m_1 = 8 / 4 = 2 \, m/s^2$. Statement $C$ is correct.Slipping starts when the acceleration of $Q$ exceeds $2 \, m/s^2$. For the whole system,$F - f_{2,max} = (m_1 + m_2) a$. At $a = 2 \, m/s^2$,$F - 9 = (9) 	imes 2 \Rightarrow F = 27 \, N$. Statement $D$ is incorrect.

The coefficient of friction between $4 \, kg$ and $5 \, kg$ blocks is $0.2$ and between $5 \, kg$ block and ground is $0.1$ respectively. Choose the correct statements.

Similar Questions

Two blocks of masses $M$ and $m$ are placed on one another on a smooth horizontal surface as shown in the figure. The force $F$ is acting on the mass $M$ horizontally during time interval $t$. Assuming no relative sliding between the blocks,the work done by friction on the blocks is .......... .

If $100 \,N$ force is applied to $10 \,kg$ block as shown in the diagram,the acceleration of $40 \,kg$ slab is (in $\,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module