A) \[\frac{ma_{1}^{2}}{4{{t}_{1}}}(t_{2}^{3}-t_{1}^{3})\]
B) \[\frac{ma_{1}^{2}}{8t_{1}^{2}}(t_{2}^{4}-t_{1}^{4})\]
C) \[\frac{ma_{1}^{2}}{4t_{1}^{2}}(t_{2}^{4}-t_{1}^{4})\]
D) \[\frac{m{{a}_{1}}}{4{{t}_{1}}}(t_{2}^{2}-t_{1}^{2})\]
Correct Answer: B
Solution :
\[W={{K}_{f}}-{{K}_{i}}=\frac{1}{2}\,(v_{f}^{2}\,-v_{i}^{2})\] Hence \[{{v}_{i}}=\frac{1}{2}{{a}_{1}}{{t}_{1}},\,\,{{v}_{f}}\,=\frac{1}{2}\,{{a}_{2}}{{t}_{2}}\] \[=\frac{1}{2}\,m\left[ {{\left( \frac{1}{2}{{a}_{2}}{{t}_{2}} \right)}^{2}}-\,{{\left( \frac{1}{2}{{a}_{1}}{{t}_{1}} \right)}^{2}} \right]\] \[=\frac{1}{8}\,m\left[ \frac{t_{2}^{4}}{t_{1}^{2}}a_{1}^{2}\,-a_{1}^{2}t_{1}^{2} \right]\,\,\left( \text{since,}\,{{\text{a}}_{2}}\,=\frac{{{t}_{2}}}{{{t}_{1}}}{{a}_{1}} \right)\] \[=\frac{ma_{1}^{2}}{8t_{1}^{2}}\,(t_{2}^{4}\,-t_{1}^{4})\]
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