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 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 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 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 5 kg is initially at rest on a rough horizontal surface. A force of 45 N acts over a distance of 2 m . The force of friction acting on the block is 25 N . The final kinetic energy of the block 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 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.2.33 J B.8.67 J C.5.33 J D.6.67 J

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 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 1.5 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