Question

The block of mass '$m$' initially at $x=0$ is acted upon by a horizontal force $F=a-bx$ as shown in the figure. The coefficient of friction between the surfaces of contact is $\mu$ . The net work done on the block is zero if the block travels a distance of ________.

• A. $(a-\mu mg)/2b$
• B. $(a-\mu mg)/b$
• C. $2(a-\mu mg)/b$
• D. $(a-2\mu mg)/b$

Answer 1

Answer: (C)

$f=a-bx$
$f_{n_{ef}}=a-bx-f_s$
For block to move
at $x=0$
$\mathrm{a}-{\mathrm{b}}_{\mathrm{x}}>{\left({\mathrm{f}}_{\mathrm{s}}\right)}_{\max }=\mu \mathrm{mg}$
$a>\mu mg$,
So we are assuming motion starts at $x=0$
$w=\int (a-bx-\mu mg)dx$
$o=ax-\frac{b{x}^2}{2}-\mu mgx$
$\Rightarrow x\left(a-\frac{bx}{2}-\mu mg\right)=0$
$x=0$
$a-\frac{bx}{2}-\mu mg=0$
$\frac{bx}{2}=(a-\mu mg)$
$x=\frac{2}{b}(a-\mu mg)$