Solve the following :
Question:

A force $F=a+b x$ acts on a particle in the $x$-direction, where $a$ and $b$ are constants. Find the work done by this force during a displacement from $x=0$ to $x=d$.

Solution:

Work done, $W={ }_{0}^{d d}$

$F d x(F=a+b x)$

$W=\left(a+\frac{1}{2} b d\right) d$

Administrator

Leave a comment

Please enter comment.
Please enter your name.