Question

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

• A. $30J$
• B. $5J$
• C. $25J$
• D. $20J$

Answer 1

The correct option is C $25J$

$W={\int }_{y_1}^{y_2}Fdy={\int }_0^1(20+10y)dy=20{\left(y\right]}_0^1+10{\left(\frac{y^2}{2}\right]}_0^1=25J$