Start Free Mock LoginAlready registered? Login here

Physics - NLM Question with Solution | TestHub

PhysicsNLMFBD & EquilibriumMedium2 minQB

PhysicsMediumnumerical

The value of the friction coefficient acting between the blocks of 1kg and 3kg, so that all the blocks move with same acceleration is $μ$ . Find the value of $6 μ$ . Consider table as smooth.

Answer:

4.00

Solution:

Common acceleration of system $\frac{3 g - 1 g}{3 + 1 + 3 + 1} = \frac{g}{4} = 2.5 m / s^{2}$

For 3 kg block $3 g - T = 3 \times \frac{g}{4} \Rightarrow T = \frac{9}{4} g = 22.5 N$ ,

For 1 kg block $T - f_{r} = 1 \times a \Rightarrow f_{r} = 22.5 - 2.5 = 20 N$

But $μ (3 g) = f_{r} = 20 \Rightarrow μ = \frac{20}{30} = \frac{2}{3} \Rightarrow 6 μ = 4$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:FBD & Equilibrium

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...