An inclined plane is bent in such a way that the vertical cross-section is given by y=`(x^2)/(4)` where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction `mu` = 0.5, the maximum height in cm at which a stationary block will not slip downward is______cm.

A block of mass 'm' is placed on a rough surface with a vertical cross section of y =(x^(3))/(6) . If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is .

A block of mass is placed on a surface with a vertical cross section given by y=x^3/6 . If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is:

A block of mass m is at rest on an incline having an angle of inclination theta with the horizontal. The incline is now being rotated with constant angular velocity omega about an axis passing through the perpendicular side. If the surface of the incline is rough having coefficient of friction as mu , then find the distance 'x' aong the incline so that block just remains stationary with restpect to incline and is on the verge of slipping. [ Take 'x' from the top of the incline as shown in the figure.]

A small mass slides down an inclined plane of inclination theta the horizontal the coefficient of friction is m = mu_(0)x , where x is the distance by the mass before it stops is .

A small back starts slipping down from a point B on an inclined plane AB, which is making an angle theta with the remaining section CA is rough with a coefficient of friction mu . It is found that bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by mu = K "tan" theta The value of K is ________.

A cannon of mass 2m located at the base of an inclined plane shoots a shell of mass m in horizontal direction with velocity v_(0) The angle of inclination of plane is 45^(@) and the coefficient of friction between the cannon and the plane is 0.5 The height to which cannon ascends the plane as a result of recoil is :

The trajectory of a projectile in a vertical plane is y = ax - bx^2 , where a and b are constant and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

The trajectory of a projectile in a vertical plane is y=ax-bx^(2) , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection form the horizontal are:

In the adjacent figure, x-axis has been taken down the inclined plane. The coefficient of friction varies with x as mu = kx , where k = tan theta . A block is released at O The maximum velocity of block will be :