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JEE Main 202125 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Evening Shift
Subject: Physics
Chapter: Work, Power and Energy

JEE Main 2021 · 25 July · Evening Shift · Physics · Work, Power and Energy

A force of F = ( 5y + 20 ) ^ j N acts on a particle. The work done...

A force of

F = ( 5y + 20 ) \overset{^}{j}N

acts on a particle. The work done by this force when the particle is moved from

y = 0m

to

y = 10m

is ________

J

.

Correct answer

0

Step-by-step solution

$F = ( 5y + 20 ) \overset{^}{j}$

$\omega = ∫ Fdy = {∫}_{0}^{10} ( 5y + 20 ) dy$

$= {\left(\dfrac{5{y}^{2}}{2} + 20y\right)}_{0}^{10}$

$= \dfrac{5}{2} \times 100 + 20 \times 10$

$= 250 + 200 = 450J$

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