A force `vecF=(5hati+3hatj)` Newton is applied over a particle which displaces it from its origin to the point `vecr=(2hati-1hatj)` metres. The work done on the particle is

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 F = (5 hati + hatj) newtons is is applied over a partical which displaced it from its origin to the point r = (2 hati - hatj) metres . The work done on the partical is.

A particle is acted upon by constant forces 4hati +hatj - 3hatk and 3hati + hatj -hatk which displace it from a point hati + 2hatj + 3hatk to the point 5hati + 4hatj + hatk . The work done in standard units by the forces is given by:

A uniform force of (3hati+hatj) N acts on a particle of mass 2kg. Hence, the particle is displaced from position (2hati+hatk) m to position (4hati+3hatj-hatk) m. The work done by the force on the particle is

A force vec(F) = 4 hati + hatj + 3hatk newton acts on a particle and displaces it through displacement vec(S) = 11 hati + 11 hatj +15 hatj metre. Calculate the work done by the force.

A force (3hati+4hatj) newton acts on a boby and displaces it by (3hati+4hatj) metre. The work done by the force is

Forces of magnitudes 5 and 3 units acting in the directions 6hati + 2hatj + 3hatk and 3 hati - hati +6hatk respectively act on a particle which is displaced from the point ( 2,2,-1) to ( 4,3,1) . The work done by the forces, is