A block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an axis is applied to the block. The force is given by `vecF=(1-x^(2))hatiN`, where x is in meters and the initial position of the block is `x=0` the maximum kinetic energy of the block is `(2)/(n)J` in between `x=0` ad `x=2`m find the value of `n`.

A 1.5-kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by vecF=(4-x^2)veciN , where x is in meter and the initial position of the block is x=0 . The maximum kinetic energy of the block between x=0 and x=2.0m is

A 1.5-kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of x-axis is applied to the block. The force is given by vecF=(4-x^2)veciN , where x is in meter and the initial position of the block is x=0 . The maximum positive displacement x is

A 1.0 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x axis is applied to the block. The force is given by vec(F)(x)= (2.5-x^(2)) hat(i)N , where x is in meters and the initial position of the block is x=0 . (a) What is the kinetic energy of the block as it passes through x=2.0m ? (b) What is the maximum kinetic energy of the block between x=0 and x=2.0 m ?

A 1.5 kg block is initially at rest on a horizontal frictionless surface. A horizontal force vec F=(4-x^(2))hat i is applied on the block. Initial position of the block is at x = 0 . The maximum kinetic enery of the block between x = 0 and x = 2m is

A block of mass 2 kg, initially at rest on a horizontal floor, is now acted upon by a horizontal force of 10 N . The coefficient of friction between the block and the floor is 0.2. If g = 10 m//s^(2) , then :

A block of mass 2 kg rests on a horizontal surface. If a horizontal force of 5 N is applied on the block, the frictional force on it is ("take", u_(k)=0.4, mu_(s)=0.5)

A block of mass 2 kg is initially at rest on a horizontal frictionless surface. A horizontal froce overlineF = (9 -x^(2)) overlinei newton acts on it, when the block is at x = 0 . The maximum kinetic energy of the block between x= 0 m and x = 3 m in joule is Conservation of mechanical energ

A block of mass 4 kg is initially at rest on a horizontal frictionless surface. A force vec(F) = (2x+3x^(2))hat(i) N acts horizontally on it. The maximum kinetic energy of the block between x = 2m and x=4m in joules is

A block of mass 4 kg is initially at rest on a horizontal frixtionless surface. A horizontal force vec(F) = (3+x) hat(i) newtons acts on it, when the block is at x = 0 . The maximum kinetic energy of the block between x = 0 and x = 2 m is

A block of mass 2 kg si kept at rest at x = 0 , on a rough horizontal surface with coefficient of a rough horizontal surface with coefficient of friction 0.2 . A position dependent horizontal force given by F = x^(2)+2x+5 (where x is in meter and F is in Newton), starts acting on the block at x = 0 . The kenetic energy of block at x = 3 is :