The only force acting on a block is along x-axis is given by `P=-((4)/(x^(2)+2))N`, When the block moves from x= -2 m to x = 4 m, the change in kinetic energy of block is-

Force acting on a block moving along x-axis is given by F = - ((4)/(x^(2)+2))N The block is displaced from x = - 2m to x = + 4m , the work done will be

Force acting on a block moving along x-axis is given by F = - ((4)/(x^(2)+2))N The block is displaced from x = - 2m to x = + 4m , the work done will be

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 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 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 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. 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