Most Important Selected Qs for JEE AdvancedPhysicsWork, Power and Energy
The potential energy of a particle of mass 0.1 kg , moving along the x -axis, is given by U =5 x ( x -4) J , where x is in meters. It can be concluded that
Options
- AThe particle is acted upon by a variable force.
- BThe minimum potential energy during motion is -20 J
- CThe speed of the particle is maximum at x =2 ~m .
- DThe period of oscillation of the particle is ( / 5 ) sec.
Correct answer
D. The period of oscillation of the particle is ( / 5 ) sec.
Step-by-step solution
F=- d U d x =-5(2 x-4) At mean position F =0 x =2 ~m aligned & U_ =-20 ~J & a =-50 2( x -2) & =10 rad / sec & T = / 5 sec aligned