An army vehicle of mass $1000\, kg$ is moving with a velocity of $10 \,m/s$ and is acted upon by a forward force of $1000\, N$ due to the engine and a retarding force of $500 \,N$ due to friction. ........... $m/s$ will be its velocity after $10\, s$

A hockey player is moving northward and suddenly turns westward with the same speed to avoid an oopponent. The force that acts on the player is

Maximum value of static friction is called

Explain -“Static friction force opposes impending motion”.

As shown in the figure a block of mass $10\,kg$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^{\circ}$, with horizontal. For $\mu_{ s }=0.25$, the block will just start to move for the value of $F..........\,N$ : $\left[\right.$ Given $\left.g =10\,ms ^{-2}\right]$

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then

$Image$

$(A)$ $\mu_1=0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{2}$

$(B)$ $\mu_1 \neq 0 \mu_2=0$ and $N_1 \tan \theta=\frac{m g}{2}$

$(C)$ $\mu_1 \neq 0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{1+\mu_1 \mu_2}$

$(D)$ $\mu_1=0 \mu_2 \neq 0$ and $N _1 \tan \theta=\frac{ mg }{2}$