JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 23)
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
450
Explanation
F = (5y + 20)$$\widehat j$$
$$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} } $$
$$ = \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$$
$$ = {5 \over 2} \times 100 + 20 \times 10$$
$$ = 250 + 200 = 450$$ J
$$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} } $$
$$ = \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$$
$$ = {5 \over 2} \times 100 + 20 \times 10$$
$$ = 250 + 200 = 450$$ J
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