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 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 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 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 1.0 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 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 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 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 unknown mass is at rest on a rough, horizontal surface. A horizontal force F is applied to the block. The graph in the figure shows the acceleration of the block with respect to the applied force. The mass of the block is

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)