JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 9)
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 :
50.0 J
12.5 J
25.0 J
0 J
Explanation
$$W = \int {\overrightarrow F \,.\,d\overrightarrow r } $$
$$ = \int\limits_1^2 {4xdx + \int\limits_2^3 {3{y^2}dy} } $$
$$ = [2{x^2}]_1^2 + [{y^3}]_2^3$$
$$ = 2 \times 3 + (27 - 8)$$
$$ = 25$$ J
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