Consider the given system in which two blocks connected by ideal string are in a lift, which is moving upward with an acceleration of $\frac{g}{2} m / s^{2}$ as shown in figure. Mass of the blocks A & B are 0.1 kg each & friction coefficient between floor & block A is $μ = 1 / 3. (g = 10 m / s^{2})$ . All units are in S.I. all are at rest at t=0 (All strings and pulleys are ideal)
Work done by tension on block B from ground frame of reference from t=0 sec. to t=3 sec. is ……J.

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answer is D.

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Detailed Solution

ANS-D

$m_{A} = m_{B} = 0.1 k g μ = \frac{1}{3} g = 10 m s^{- 1} N = \frac{3}{2} f_{\max} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} N \frac{3}{2} - T = 0.1 a^{1} T - f = 0.1 a^{1} \frac{3}{2} - \frac{1}{2} = 0.2 a^{1} \frac{1}{0.2} = a^{1} \Rightarrow a^{1} = 5 m s^{- 2} \frac{3}{2} - T = 0.5 T = 1 N a^{1} = 5 m s^{- 2} a_{b x} = 5 m s^{- 2} x = \frac{1}{2} (5) \times (\frac{4}{2}) x = 2 m W_{b y s t r i n g} = 2 J T = 1 N a_{B y} = a_{b t y} + a_{t y} = 5 ↓ + 5 ↑ = 0 S = 0$
Work done is zero
3.ANS-A
4.ANS- C
SOL- If block is observed from inclined plane then FBD is shown
$N = m g \cos θ + m w^{2} R \sin θ - (A) f_{r} + m w^{2} R \cos θ = m g \sin θ μ N = m g \sin θ - m w^{2} \sin θ - (B)$
Solving A&B for w
$w = \sqrt{\frac{g}{R} (\frac{\tan θ - μ}{1 + μ \tan θ})} = 3.2 \frac{r a d}{\sec}$

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