Calculate the stopping distance of a vehicle of mass $M$ moving with velocity $v$ on a straight road. (Where $\mu$ is the coefficient of friction between the tires and the road.)

$A$ body of mass $10 \,kg$ is kept on a rough horizontal surface with a coefficient of friction of $0.3$. If a horizontal force of $50 \,N$ is applied to the body, then the acceleration of the body is (Acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-2}$)

$A$ car is moving along a straight horizontal road with a speed $v_0$. If the coefficient of friction between the tyres and the road is $\mu$,the shortest distance in which the car can be stopped is

$A$ body of weight $64 \ N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.8$ and $0.6$ respectively,then the acceleration of the body will be

$A$ block of mass $0.5 \ kg$ is at rest on a horizontal table. The coefficient of kinetic friction between the table and the block is $0.2$. If a horizontal force of $5 \ N$ is applied on the block,the kinetic energy of the block in a time of $4 \ s$ is (Acceleration due to gravity $= 10 \ m/s^2$). (in $J$)

$A$ wooden block of mass $m$ resting on a rough surface (coefficient of friction $\mu$) is pulled by a force $F$ at an angle $\theta$ with the horizontal,as shown in the figure. The acceleration of the block moving horizontally is: