A force of F = (5y + 20)$\widehat j$ 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.
Answer (integer)
450
Solution
F = (5y + 20)$\widehat j$<br><br>$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} }$<br><br>$= \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$<br><br>$= {5 \over 2} \times 100 + 20 \times 10$<br><br>$= 250 + 200 = 450$ J
About this question
Subject: Physics · Chapter: Work, Energy and Power · Topic: Work Done by a Force
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