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  • A block starts moving up an inclined plane of inclination  with an initial velocity of 

A block starts moving up an inclined plane of inclination 30^{\circ} with an initial velocity of \nu _{0}. It comes back to its initial position with velocity \frac{\nu _{0}}{2}. The value of the coefficient of kinetic friction between the block and the inclined plane is close to \frac{I}{1000}, The nearest integer to I is _______.
Option: 1 346
Option: 2 12345
Option: 3 7689
Option: 4 12341

Answers (1)

best_answer

Let the length of the inclined plane as S (as shown in the figure)

 

For upward motion 

\begin{aligned} &\text { from work energy theorem }\\ &W_{f}=k_{f}-k_{i} \end{aligned}

So

\frac{1}{2} m v_{0}^{2}=\left(m g \frac{S}{2}+\mu m g \frac{\sqrt{3}}{2} S\right)....(1)

 

For downward motion 

\begin{aligned} &\text { from work energy theorem }\\ &W_{f}=k_{f}-k_{i} \end{aligned}

\frac{1}{2} \frac{m v_{0}^{2}}{4}=\left(m g \frac{S}{2}-\mu m g \frac{\sqrt{3}}{2} \cdot S\right).......(2)

From equation (1) and (2) 

we get

\begin{array}{l} 4=\frac{(1+\mu \sqrt{3})}{(1-\mu \sqrt{3})} \\ \\ \Rightarrow 4-4 \mu \sqrt{3}=1+\mu \sqrt{3} \\ \Rightarrow 3=5 \mu \sqrt{3} \\ \\ \therefore \mu=\frac{\sqrt{3}}{5} \approx \frac{346.41}{1000} \\ I=346 \end{array}

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avinash.dongre

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