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QJEEM21Ph1d4f: JEE Main 2021 Physics - Class 11 JEEM21Ph1d4f Solution: A force of F = (5y + 20) N acts on a particle. The work done by …

A force of F = (5y + 20) j^\widehat j N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J.

Step-by-Step Verified Solution

F = (5y + 20) j^\widehat j

W=∫Fdy=∫010(5y+20)dyW = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} }

=(5y22+20y)010= \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}

=52×100+20×10= {5 \over 2} \times 100 + 20 \times 10

=250+200=450= 250 + 200 = 450 J

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