Most Important Selected Qs for JEE AdvancedPhysicsWork, Power and Energy
A particle of mass 2 kg moves on a smooth horizontal plane under the action of a single force F =(30 i +40 j ) N . Under this force it is displaced from (0,0) to (100 ~m , 100 ~m ) . If the initial speed of the particle is 3000 ~m / s then find out its final speed in m / s .
Correct answer
100
Step-by-step solution
aligned & W = KE = 1 2 mv ₂^2- 1 2 mv ₁^2 Here W = F d =(30 i +40 j ) (100 i +100 j )=3000+4000=7000 & 7000= 1 2 2 v ₂^2- 1 2 2 3000 v ₂^2=10000 v ₂=100 aligned