A) 12.50 N-m
B) 18.75 N-m
C) 25.00 N-m
D) 6.25 N-m
Correct Answer: B
Solution :
Magnitude of work done \[|{{W}_{1}}|\,=\,\frac{1}{2}\,k\times x_{1}^{2}\] = Magnitude of elastic potential energy stored in spring block system. \[=\frac{1}{2}\,\times 5\times {{10}^{3}}\times {{(5\times {{10}^{-2}})}^{2}}\] \[\Rightarrow \] \[|{{U}_{1}}|\,=\,6.25\,J\] Now trually \[|{{W}_{2}}|\,=\,\frac{1}{2}\,k{{({{x}_{1}}+{{x}_{2}})}^{2}}\] \[=\frac{1}{2}\,\times 5\times {{10}^{3}}\,{{(5\times {{10}^{-2}}+5\times {{10}^{-2}})}^{2}}\] \[|{{U}_{2}}|=25\,J\] Net work done \[=\,|{{W}_{2}}|\,-\,|{{W}_{1}}|=|{{U}_{2}}|\,-\,|{{U}_{1}}|\] \[=25-6.25\] \[=18.75\,\,J=18.75\,\,N-m\]
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