Question
A body of $5 \mathrm{~kg}$ weight kept on a rough inclined plane of angle $30^{\circ}$ starts sliding with a constant velocity. Then the coefficient of friction is (assume $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )(a) $1 / \sqrt{3}$(b) $2 / \sqrt{3}$(c) $\sqrt{3}$(d) $2 \sqrt{3}$
Step 1
Here, the mass is $5 \mathrm{~kg}$ and gravity is $10 \mathrm{~m} / \mathrm{s}^{2}$. So, the weight of the body is $5 \times 10 = 50 \mathrm{~N}$. Show more…
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A block of mass $5.0 \mathrm{~kg}$ slides down from the top of an inclined plane of length $3 \mathrm{~m}$. The first $1 \mathrm{~m}$ of the plane is smooth and the next $2 \mathrm{~m}$ is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at $30^{\circ}$ with the horizontal (Fig. $7.449$ ), find the coefficient of friction on the rough portion. a. $\frac{2}{\sqrt{3}}$ b. $\frac{\sqrt{3}}{2}$ c. $\frac{\sqrt{3}}{4}$ d. $\frac{\sqrt{3}}{5}$
A block of mass $3 \mathrm{~kg}$ slides down an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of kinetic friction between the block and surface is $\frac{1}{2 \sqrt{3}}$. The forces that must be applied to the block for it (i) to move down the plane without any acceleration (ii) to move up the plane without any acceleration are respectively $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) $7.5 \mathrm{~N}, 22.5 \mathrm{~N}$ (b) $40 \mathrm{~N}, 25 \mathrm{~N}$ (c) $20 \mathrm{~N}, 50 \mathrm{~N}$ (d) $15 \mathrm{~N}, 21 \mathrm{~N}$
A body of mass $1 \mathrm{~kg}$ has velocity $1 \mathrm{~m} \mathrm{~s}^{-1}$, up an inclined plane of angle of $30^{\circ}$ to the horizontal. The friction coefficient is $\frac{1}{\sqrt{3}} .$ The distance the body travels before stopping is $\left(g=10 \mathrm{~m} \mathrm{~s}^{2}\right)$ (a) $5 \mathrm{~cm}$ (b) $7.5 \mathrm{~cm}$ (c) $10 \mathrm{~cm}$ (d) $6.7 \mathrm{~cm}$
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