Question

A particle of mass 'm' is moving along X-axis under a force F = k√x. Work done for the motion of particle from x = 4m to x = 9 m is nearly

I need full explanation plz​

Answer 1

Answer:

8.6k

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Answer 2

Final Answer: Work done for the motion of a particle from x = 4m to x = 9m is 12.67kJoule.

Explanation: Given that,

F = k√x

Work done for the motion of particle from x = 4m to x = 9m is

$W = \int\limits^a_b {F} \, dx \\W = \int\limits^9_4 {k\sqrt{x} } \, dx \\W = k[\frac{x^\frac{3}{2} }{\frac{3}{2} } ]^9_4\\W = k\frac{2}{3}[x^\frac{3}{2} ]^9_4\\W = \frac{2}{3}k[9^\frac{3}{2} - 4^\frac{3}{2}\\W = \frac{2}{3}k[3^2^(^3^/^2^) - 2^2^(^3^/^2^) \\W = \frac{2}{3}k[3^3 - 2^3]\\W = \frac{2}{3}k[27 - 8]\\W = k\frac{2}{3}(19)\\W = 12.66k\\W = 12.67kJoule.$