A force vecF=(3xN)hati+(4N)hatj, with x ...

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?

Similar Questions

Explore conceptually related problems

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 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

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 ?

A force vecF=(5hati+3hatj+2hatk)N is applied over a particle which displaces it from its origin to the point vecr=(2hati-hatj) m. The work done on the particle in joules 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 vec F=(6 hati -8 hatj)N , acts on a particle and displaces it over 4 m along the X-axis and 6 m along the Y-axis. The work done during the total displacement is

A force vecF=(4hati+5hatj)N acts on a particle moving in XY plane. Starting from origin, the particle first goes along x-axis to the point (5.0) m and then parallel to the Y-axis to the point (5,4 m. The total work done by the force on the particle 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?

Similar Questions

Recommended Questions