SATHEE: Laws of Motion - Result Question 23

Laws of Motion - Result Question 23

23. A system consists of three masses $m_{1}, m_{2}$ and $m_{3}$ connected by a string passing over a pulley P. The mass $m_{1}$ hangs freely and $m_{2}$ and $m_{3}$ are on a rough horizontal table (the coefficient of friction $= μ$ ). The pulley is frictionless and of negligible mass. The downward acceleration of mass $m_{1}$ is : $(.$ Assume $. m_{1} = m_{2} = m_{3} = m)$

[2014]

(a) $\frac{g (1 - g μ)}{g}$

(b) $\frac{2 g μ}{3}$

(d) $\frac{g (1 - 2 μ)}{2}$

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Answer:

Correct Answer: 23. (c)

Solution:

$= \frac{Net force in the direction of motion}{Total mass of system}$

$\begin{aligned} = \frac{m_{1} g - μ (m_{2} + m_{3}) g}{(m_{1} + m_{2} + m_{3})} = \frac{g}{3} (1 - 2 μ) \\ (∵ m_{1} . . = m_{2} = m_{3} = m given) \end{aligned}$

Laws of Motion - Result Question 23

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Laws of Motion - Result Question 23

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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