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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A position dependent force F = (10-2x +3x$^2$ ) N acts on a small object of mass 2 kg to displace it from x = 0 to x = 3m. The work done in Joules is

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Answer: 363 J
Given a variable force $F = (10-2x +3x^2 ) N$, Work done $W = \large\int \limits_a^b$$ F dx \rightarrow W = \large\int \limits_0^3$$ 10-2x+3x^2 dx $
$\Rightarrow W = [10x - 2x^2 + 3x^3]_0^3 = 10 (3) - 2 (3^2) + 3 (3^3) = 300-18 + 81 =363J$
answered Aug 26, 2014 by balaji.thirumalai
 

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