"A particle experiences a variable force $\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$ 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, then Kinetic Energy changes by :"

Question

Accepted Answer

"

$W = \int {\overrightarrow F \,.\,d\overrightarrow r }$

$= \int\limits_1^2 {4xdx + \int\limits_2^3 {3{y^2}dy} }$

$= [2{x^2}]_1^2 + [{y^3}]_2^3$

$= 2 \times 3 + (27 - 8)$

$= 25$ J

"

A particle experiences a variable force $\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$ 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, then Kinetic Energy changes by :

Solution

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A particle experiences a variable force $\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$ 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, then Kinetic Energy changes by :

Solution

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