Home
Class 11
PHYSICS
A particle of mass 0.5 kg is displaced f...

A particle of mass `0.5 kg` is displaced from position `vec r_(1)(2,3,1)` to`vec r_(2)(4,3,2)` by applying a force of magnitude `30 N` which is acting along `(hati + hatj + hatk)`. The work done by the force is

A

`10sqrt(3)J`

B

`30sqrt(3)J`

C

30 J

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • WORK, POWER AND ENERGY

    DC PANDEY ENGLISH|Exercise A Only One Option is Correct|42 Videos
  • WORK, POWER AND ENERGY

    DC PANDEY ENGLISH|Exercise B More than One Option is Correct|26 Videos
  • WORK, ENERGY AND POWER

    DC PANDEY ENGLISH|Exercise MEDICAL ENTRACES GALLERY|33 Videos

Similar Questions

Explore conceptually related problems

Two constant forces vecF_1 and vecF_2 act on a body of mass 8kg . These forces displace the body from point P(1, -2, 3) to Q (2, 3, 7) in 2s starting from rest. Force vecF_1 is of magnitude 9N and is acting along vector (2hati-2hatj+hatk) . Work done by the force vecF_2 is

A force F=(2hati-hatj+4hatk)N displaces a particle upto d=(3hati+2hatj+hat k)m. Work done by the force is

A particle is displaced from a position 2hati-hatj+hatk (m) to another position 3hati+2hatj-2hatk (m) under the action of a force 2hati+hatj-hatk(N) . The work done by the force is

A force (3hati+4hatj) newton acts on a boby and displaces it by (2hati+3hatj) metre. The work done by the force is

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

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

A body moves from a position vec(r_(1))=(2hati-3hatj-4hatk) m to a position vec(r_(2))=(3hati-4hatj+5hatk)m under the influence of a constant force vecF=(4hati+hatj+6hatk)N . The work done by the force is :

A force vecF= (3hati+4hatj) N acts on a body and displaces it by vecS= (3hati+4hatj)m . The work done (W= vecF*vecS ) by the force is :

A body is displaced from (0,0) to (1 m, 1m) along the path x = y by a force F=(x^(2) hatj + y hati)N . The work done by this force will be

A force vecF=(2hati+3hatj+4hatk)N displaces a body from position vector vec(r_(1))=(2hati+3hatj+hatk)m to the positive vector vec(r_(2))=(hati+hatj+hatk)m . Find the work done by this force.