Work Energy Power

JEE Main level, single correct, medium

92 of 183

Question 92 of 183

A particle experiences a variable force F⃗=(4xi^+3y2j^)\vec{F}=\left(4x\hat{i}+3{y}^{2}\hat{j}\right) in a horizontal x−yx-y plane. Assume distance in meters and force is newton. If the particle moves from point (1,2)\left(1,2\right) to point (2,3)\left(2,3\right) in the x−yx-y plane, then Kinetic Energy changes by :

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Answer: (A) 25J25J

Using Work energy theorem,
Kinetic energy ∆KE=W=∫124xdx+∫233y2dy∆KE=W={\int }_{1}^{2}4xdx+{\int }_{2}^{3}3{y}^{2}dy
=[4x22]12+[3y33]23={\left[\dfrac{4{x}^{2}}{2}\right]}_{1}^{2}+{\left[\dfrac{3{y}^{3}}{3}\right]}_{2}^{3}
=[2×22−2×12]+[32−23]=6+19=\left[2\times {2}^{2}-2\times {1}^{2}\right]+\left[{3}^{2}-{2}^{3}\right]=6+19
=25=25 Joules

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