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A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic frictions are 0.8 and 0.6 respectively, then the acceleration of the body will be
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0.2 g
Given, weight of body, W = 64 N
Coefficient of static friction, $\mu_s=0.8$
Coefficient of dynamic friction, $\mu_d=0.6$
Force required to move the body from rest
= Coefficient of static friction * Weight of the body
$\begin{aligned} & =\mu_s \times m g \\ & =0.8 \times 64...(i)\end{aligned}$
As weight, W = 64 N
$\therefore$ Mass of the body, $m=\frac{64}{g}$
As we know, F = ma
From Eqs. (i) and (ii), we get
$\begin{aligned} F & =m g\left(\mu_s-\mu_d\right) \\ m a & =64(0.8-0.6) \\ \Rightarrow \quad \frac{64}{g} \times a & =64(0.2) \\ \Rightarrow \quad a & =0.2 \mathrm{~g}\end{aligned}$
Coefficient of static friction, $\mu_s=0.8$
Coefficient of dynamic friction, $\mu_d=0.6$
Force required to move the body from rest
= Coefficient of static friction * Weight of the body
$\begin{aligned} & =\mu_s \times m g \\ & =0.8 \times 64...(i)\end{aligned}$
As weight, W = 64 N
$\therefore$ Mass of the body, $m=\frac{64}{g}$
As we know, F = ma
From Eqs. (i) and (ii), we get
$\begin{aligned} F & =m g\left(\mu_s-\mu_d\right) \\ m a & =64(0.8-0.6) \\ \Rightarrow \quad \frac{64}{g} \times a & =64(0.2) \\ \Rightarrow \quad a & =0.2 \mathrm{~g}\end{aligned}$
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