KVPY Sample Paper KVPY Stream-SX Model Paper-21

  • question_answer
    A mass m in three-dimensional space is subjected to three forces: \[\theta \] \[{{F}_{2}}\] and \[{{F}_{3}},\] \[{{F}_{1}}\] and \[{{F}_{2}}\] have the same magnitude, with \[{{F}_{1}}\] in the positive-x direction, and \[{{F}_{2}}\] in the positive-y direction. If the mass has an acceleration of 0, which of the following statements is false?

    A) The magnitude of \[{{F}_{3}}\] is the same as that of \[{{F}_{1}}\]

    B) The object is in equilibrium and could be stationary

    C) \[{{F}_{3}}\] lies in the x-y plane

    D) The object is in equilibrium and could be moving

    Correct Answer: A

    Solution :

    By definition, the object is in equilibrium, either static (unmoving) or dynamic (moving with a constant velocity). If the object has acceleration a=0, the net force acting on the mass must be 0 as well:
    With force \[{{F}_{1}}\]and \[{{F}_{2}}\] in the x-y plane, the force that will counteract them must lie in the x-y plane as well, as shown. The magnitude of that force \[{{F}_{3}}\] is equal to the vector sum of \[{{F}_{1}}\] and \[{{F}_{2}}\] and can be calculated as follows:
    \[\left| \,{{F}_{3}} \right|=\sqrt{F_{3-x}^{2}+F_{3-y}^{2}}\]\[=\sqrt{F_{1}^{2}+F_{2}^{2}}={{F}_{1}}\sqrt{2}\]
    both graphically and analytically, we can see that the magnitude of \[{{F}_{3}}\] is not the same as that of \[{{F}_{1}}.\]

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