[" Q.6The potential energy of a particle of mass "0.1kg," moving along "x" -axis,is given by "U=5x(x-4)J" ."],[x" is in metres.It can be concluded that "]

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energy of a particle of mass 1 kg moving along x-axis given by U(x)=[(x^(2))/(2)-x]J . If total mechanical speed (in m/s):-

The potential energy of a particle of mass 1 kg moving along x-axis given by U(x)=[(x^(2))/(2)-x]J . If total mechanical speed (in m/s):-

The potential energy of a particle of mass 1 kg in motin along the x-axis is given by U = 4(1-cos2x)J Here, x is in meter. The period of small osciallationis (in second) is