Here in this case we can use work energy theorem
As per work energy theorem
Work done by all forces = Change in kinetic Energy of the object
Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.
Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy
[tex]KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2[/tex]
also we know that
[tex]I = \frac{2}{5}mR^2[/tex]
[tex] w= \frac{v}{R}[/tex]
Now kinetic energy is given by
[tex]KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2[/tex]
[tex]KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2[/tex]
[tex]KE = \frac{7}{10}mR^2w^2 [/tex]
[tex]KE = \frac{7}{10}*150*(0.200)^2(50)^2[/tex]
[tex]KE = 10500 J[/tex]
Now by work energy theorem
Work done = 10500 - 0 = 10500 J
So in the above case work done on sphere is 10500 J
