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

Question

A particle experiences a variable force F=(4xi+3y^2j) in a horizontal x-y plane. 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, the Kinetic Energy changes by

Accepted Answer

Correct answer: C. (c) F=4xi+3y^2j WD= KE W= F .( dxi+ dyj) = 4 xdx+ 3y^2 dx = (2x^2 )_1^2+ (y^3 )_2^3 = (8-2) + (27-8) = 6 + 19 = 25J

$A particle experiences a variable force F = (4 x i + 3 y^{2} j) in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, the Kinetic Energy changes by$

Solution

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