Sign convention: +x = block to the right, +y = hanging mass downward
Given
mblock=2.00kg,mhang=7.00kg,μ=0.00,g=9.81m/s2Newton's 2nd law (system)
Fnet=mtotal⋅a mtotal=mblock+mhang=2.00+7.00=9.00kg Forces
Fpull=mhang⋅g=7.00×9.81=68.67N Ffric=μN=0(friction disabled) Fnet=Fpull−Ffric=68.67−0.00=68.67N Result
a=mtotalFnet=9.0068.67=7.63m/s2 t=0.00s