A force `vecF=(3xy-5z)hatj+4zhatk` is applied on a particle. The work done by the force when the particle moves from point `(0, 0, 0)` to point `(2, 4, 0)` as shown in figure.

A force vecF = ( 3hati + 4hatj) N is acting on a particle. Work done by the force is zero, when particle is moving on the line 2y + Px = 2 , the value of P is

A force (5y + 20) hatj displaces a particle from y=0 to y=10. Find work done by force

Force acting on a particle varies with x as shown in figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.

A force vecF = (2xhati+ 3hatj + 3z^2hatk) is acting on the particle of mass 2kg which is displaces from point A(2,3,0) to point B(3,2,1) . the work done by the force on the particle will be

A force vecF = (2xhati+ 3hatj + 3z^2hatk) is acting on the particle of mass 2kg which is displaces from point A(2,3,0) to point B(3,2,1) . the work done by the force on the particle will be

A force of F=(0.5x+12)N acts on a particle. If x is in metre, calculate the work done by the force during the displacement of the particle from x = 0 to x = 4 m

A force vecF =(5hati+3hatj+2hatk)N acts on a particle , and the particle moves from the origin to a point vecr=(2hati-hatj) m. what will be the work done on the particle ?

The velocity of a particle of mass 10^(-2)kg is given as vecv=(2xhatu+3yhatj+4zhatk) m/sec (where x, y and z are in meter). The amount of work done by applied force in displacing a particle from (1m, 2m, 0) to (-1m, 3m, 1m) is …..