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 vec F = (4 - x ^2) hati N , where x is in metre and the initial position of the block is x = 0. The maximum kinetic energy of the block between x = 0 and x = 2.0 m 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 1.5 kg 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 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 1.0 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 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 :

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 :

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 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