Question

45. The PE of a 2 kg particle, free to move along
x-axis is given by V(x)= *-* J. The total
13 2
mechanical energy of the particle is 4 J. Maximum
speed (in ms-') is​

Answer 1

Answer:

The answer will be 2 m/s

Explanation:

According to the problem the potential energy of a particle is given

Now we know that the velocity of the particle s maximum when the kinetic energy of the particle is maximum

Therefore the minimum potential energy at x = 0 is 0 J

Now it is given that the Kinetic energy(max) + potential energy(min) = 4

                                     =>Kinetic energy(max) = 4 j

Therefore 1/2 mv^2 = 4

          => 1/2 x 2 x v^2 = 4

        => v^2 = 4 => v = 2 m/s