Question

3 particles each of mass 100gm on vertices of triangle of side-20cm,workdone in increasing side 40cm

Answer 1

Answer:

Work done= Final gravitational potential -Initial gravitational potential

=vf−vi=3[(−Gmmrf)−(−Gmmri)]=vf−vi=3[(−Gmmrf)−(−Gmmri)]

m=0.1rf=0.4mri=0.2mm=0.1rf=0.4mri=0.2m

Substituting these values we get

W=3[−Gmm(1.4−1.2)]W=3[−Gmm(1.4−1.2)]

W=5.0×10−12JW=5.0×10−12J