InterviewSolution
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InterviewSolution
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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. The force is given by `vecF=(4-x^2)veciN`, 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=2.0m` isA. (a) `2.33J`B. (b) `8.67J`C. (c) `5.33J`D. (d) `6.67J` |
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Answer» Correct Answer - C From work-energy theorem, kinetic energy of the block at `x=x is K=underset0oversetx int4-x^2dx=4x-x^3/3` For K to be minimum, `(dK)/(dx)=0` or `4-x^2=0` or `x=+-2m` At `x=+2m`, `(d^2K)/(dx^2)` is negative, i.e., kinetic energy K is maximum and `K_(max)=4xx2-((2)^3)/(3) ~=5.33J` |
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