Work Power Energy Question 8
Question 8 - 01 February - Shift 2
A force $F=(5+3 y^{2})$ acts on a particle in the $y-$ direction, where $F$ is newton and $y$ is in meter. The work done by the force during a displacement from $y=2 m$ to $y=5 m$ is
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Answer: (132)
Solution:
Formula: Work Done By A Variable Force
$F=5+3 y^{2}$
$W=\int_2^{5}(5+3 y^{2}) dy$
$=[5 y+\frac{3 y^{3}}{3}]_2^{5}$
$=132 J$
