Q94 / 180 · NEET 2019 (All India) · Physics

A force $F = 20 + 10 y$ acts on a particle in $y$ -direction where $F$ is in Newton and $y$ is in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is

A
$30 J$
B
$5 J$
Correct answer
C
$25 J$
D
$20 J$

Solution

Work done is calculated by taking the dot product between the force and the displacement.
$W = \int F \cdot d y$
$= \int_{0}^{1} 20 d y + \int_{0}^{1} 10 y d y$
$= {[20 y]}_{0}^{1} + 5 {[y^{2}]}_{0}^{1}$
$= 20 + 5 = 25 J$

Hence, the correct answer is option: 'C'.

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A force F=20+10y acts on a particle in y-direction where F is in Newton and y is in meter. Work done by this force to move the particle from y=0 to y=1 m is

Solution

Relevant Questions

A force $F = 20 + 10 y$ acts on a particle in $y$ -direction where $F$ is in Newton and $y$ is in meter. Work done by this force to move the particle from $y = 0$ to $y = 1 m$ is