NEET Sample Paper NEET Sample Test Paper-21

  • question_answer
    A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of \[x-\]axis is applied to the block. he force is given by \[\vec{F}=(4-{{x}^{2}})\hat{i}\,N,\]where \[x\]is in meter and the initial position of the block is \[x=0.\]The maximum kinetic energy of the block between\[x=0\]and\[x=2.0\] m is

    A)  2.33 J                        

    B)  8.67 J

    C)    5.33 J              

    D)    6.67 J

    Correct Answer: C

    Solution :

    From work-energy theorem kinetic energy of block at \[x=x\]is; \[K=\int_{0}^{x}{(4-{{x}^{2}})}.dx\] or \[K=4x-\frac{{{x}^{3}}}{3}\] For K to be maximum \[\frac{dK}{dx}=0\] or \[4-{{x}^{2}}=0\] or \[x=\pm 2\,m\] At \[x=+2\,m,\,\frac{{{d}^{2}}K}{d{{x}^{2}}}\]is negative i.e., kinetic energy (K) is maximum. \[\therefore \]\[{{K}_{\max }}=(4)(2)-\frac{{{(2)}^{2}}}{3}=\frac{16}{3}J\,=5.33\,J\] Hence, the correction option is (c).

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