$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)

  • A
    $15$
  • B
    $30$
  • C
    $45$
  • D
    $60$

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$A$ block of mass $10\, kg$ is kept on a rough inclined plane as shown in the figure. $A$ force of $3\, N$ is applied on the block directed down the plane. The coefficient of static friction between the plane and the block is $0.6$. What should be the minimum value of force $P$ applied up the plane,such that the block does not move downward (in $, N$)? (Take $g = 10\, ms^{-2}$)

$A$ board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of the addition of a weight to one of the ends of the board when the board makes an angle $\theta$ with the horizontal. The coefficient of friction between the log and the board is:

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The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu$ is the coefficient of friction,the inclination of the plane to the horizontal is:

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

$A$ block is projected up an inclined plane of angle $\theta = 30^o$ with an initial velocity of $5 \, m/s$. It comes to rest in $0.5 \, s$. What is the coefficient of friction?

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