A force `F=(3xhati+4hatj)` Newton (where x is in metres) acts on a particle which moves from a position (2m, 3m) to (3m, 0m). Then the work done is

Force F=[3x^2hati+4hatj] N where , x in metres, acts on a particle. How much work is done on the particle as it moves from coordinates (2m,3m) to (3m, 0m)?

A particle is acted upon by a force vecF=y hati+xhatj newton. When the particle is moved from (1 m, 1 m) to (9m, 3 m) via straight path, work done by vecF is y. What is the value of x/y?

A force of (4x^(2)+3x)N acts on a particle which displaces it from x=2m to x = 3m . The work done by the force is

A force vecF=(3xN)hati+(4N)hatj , with x in meter, acts on a particle, changing only the kinetic energy of the particle. How mcuh work is done on the particle as it moves from coordinates (2m, 3m, 5m) to (3m, 0m, 6m) ? Does the speed of the particle increase, decrease, or remain the same?

A force vecF=(5hati+3hatj)N is applied over a particle which displaces it from its original position to the point vecs=S(2hati-1hatj)m . The work done on the particle is

A force vecF=xhati+y^(2)hatjN acts on a particle and the particle moves from (1,2) m to (–3,4) m . Find work done by the force vecF .

Two constat forece vecF_(1) = (2hati + 3hatj + 3hatk) newton and vecF_(2) = (5hati - 6hatj - 2hatk) newton act toghter on a particle during its displacement from the position (20hati + 15hatj) m to 8vecKm . Calculate the work done.

A particle is acted upon by only a conservative force F=(7hat(i)-6hat(j)) N (no other force is acting on the particle). Under the influence of this force particle moves from (0, 0) to (-3m, 4m) then