A particle experiences a variable force → F = (4x ^ i + 3 y ^ 2 ^ j ) in a horizontalx - yplane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x - y plane

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Quantrex Academy · Accepted Answer

Correct answer: A. 25J

Using Work energy theorem,

Kinetic energy $∆ KE = W = {∫}_{1}^{2}4x d x + {∫}_{2}^{3}3{y}^{2} d y$

$= {\left[\dfrac{4{x}^{2}}{2}\right]}_{1}^{2} + {\left[\dfrac{3{y}^{3}}{3}\right]}_{2}^{3}$

$= \left[2 \times {2}^{2} - 2 \times {1}^{2}\right] + \left[{3}^{2} - {2}^{3}\right] = 6 + 19$

$= 25$ Joules

A particle experiences a variable force → F = (4x ^ i + 3 y ^ 2 ^ j...

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