If the coefficient of friction between block & wedge is $\mu $, then the maximum horizontal acceleration of the wedge for which block will remain at rest w.r.t. the wedge is
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Two masses $M$ and $m$ are attached to a vertical axis by weightless threads of combined length $l$. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity $\omega $. If the tensions in the threads are the same during motion, the distance of $M$ from the axis is
A body of mass $10\,kg$ is moving with an initial speed of $20\,m / s$. The body stops after $5\,s$ due to friction between body and the floor. The value of the coefficient of friction is (Take acceleration due to gravity $g =10\; ms ^{-2}$)
A uniform rod of length $L$ and mass $M$ has been placed on a rough horizontal surface. The horizontal force $F$ applied on the rod is such that the rod is just in the state of rest. If the coefficient of friction varies according to the relation $\mu = Kx$ where $K$ is a $+$ ve constant. Then the tension at mid point of rod is
In figure, two blocks $M$ and $m$ are tied together with an inextensible and light string. The mass $M$ is placed on a rough horizontal surface with coefficient of friction $\mu$ and the mass $m$ is hanging vertically against a smooth vertical wall. The pulley is frictionless. When the downward acceleration of the elevator becomes equal to $g$, then
If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \,m / s$, stops after $0.5 s$, then coefficient of friction will be nearly
Two blocks $A$ and $B$ of masses $m_A = 1\,kg$ and $m_B = 3\,kg$ are kept on the table as shown in figure. The coefficient of friction between $A$ and $B$ is $0.2$ and between $B$ and the surface of the table is also $0.2.$ The maximum force $F$ that can be applied on $B$ horizontal, so that the block $A$ does not slide over the block $B$ is $...... N. [$Take $g = 10\,m/s^2 ]$
A lift is moving downwards with an acceleration equal to acceleration due to gravity. $A$ body of mass $M$ kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu $, if the lift is moving upwards with a uniform velocity, then the frictional resistance offered by the body is
A block of mass $M$ slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$. The magnitude of the contact force will be.