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AMU Medical AMU Solved Paper-2010

  • question_answer
    A variable force, given by the two- dimensional vector \[\text{F }=\left( \text{3}{{\text{x}}^{\text{2}}}\overset{\hat{\ }}{\mathop{\text{i}}}\,+\text{4}\overset{\hat{\ }}{\mathop{\text{j}}}\, \right),\] acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0)? (The coordinates are in metre)

    A)  -7 J                                       

    B)  Zero

    C)  +7 J                                      

    D) d) +19 J

    Correct Answer: C

    Solution :

                     Given, the two-dimensional force                 \[F=3{{x}^{2}}\hat{i}+4\hat{j}\]                 \[\vec{r}=x\hat{i}+y\,\hat{j}\]                 \[d\,\vec{r}=dx\,\hat{i}+dy\,\hat{j}\] Kinetic energy = Work done                 \[W=\int{F.\,\,d\,\vec{r}}\] \[=\int_{\,(2,3)}^{\,(3,0)}{(3{{x}^{2}}\hat{i}+4\,\hat{j})}.\,(dx\,\hat{i}+dy\,\hat{i})\] \[=\int_{\,(2,3)}^{\,(3,0)}{(3{{x}^{2}}}dx+4dy\,\] \[=\int_{\,(2,3)}^{\,(3,0)}{3{{x}^{2}}}dx+\int_{3}^{0}{4dy\,}\] \[=[{{x}^{3}}]_{2}^{3}+4[y]_{3}^{0}\] \[=(27-8)+4(-3)\] \[=19-12=7J\]


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