The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg is resting on a table of $\mu =0.20$ The maximum acceleration which can be produced by the cable is (Taking $g=10{\mathrm{ms}}^{-2}$)

Given that

The maximum strength of cable F = 40N

A body of mass m = 8kg has to be pulled along the table as shown.

$\text{ Coefficient of friction }{\mu }_s=0.2$

$\text{ We know }{\mathrm{f}}_{\mathrm{ms}}=\mu \mathrm{N}=0.2\times \mathrm{mg}=0.2\times 8\times 10=16\mathrm{N}$

$\text{ Maximum acceleration produced }=a$

$\text{ Maximum force }F\text{ that can be applied }=40\mathrm{N}\text{. }$

Maximum force F that can be applied = 40N.

We can have

$\begin{array}{r}\mathrm{F}-{\mathrm{f}}_{m\mathrm{s}}=\mathrm{ma}\\ 40-16=8\left(a\right)\end{array}$

$\Rightarrow 24=8a$

$\therefore a=\frac{24}{8}=3{\mathrm{ms}}^{-2}$