Question

A horizontal constant force F pulls a block of mass m placed on a horizontal surface If the coefficient of kinetic friction between the block and ground. find the power delivered by the external agent after time"t".

• A. $\frac{\mu mg}{\sqrt{1+3u^2}}$
  
• B. $\frac{\mu mg}{\sqrt{1+2u^2}}$
  
• C. $\frac{\mu mg}{\sqrt{1+u^2}}$
  
• D. $\frac{\mu 2mg}{\sqrt{1+u^2}}$
  

Answer 1

Answer: (C)

$\frac{\mu mg}{\sqrt{1+u^2}}$

Let P be the force applied to it at an angle $\theta$
From the free-body diagram "
$R+Psin\theta -mg=0$

$R=-PSin\theta +mg$

$mR=Pcos\theta$

$\mu (Psin\theta -Pcos\theta )$

$P=\frac{\mu mg}{\mu Sin\theta }+cos\theta$

Applied force should be minimum when $msin\theta +cos\theta$ is maximum.
Again,
$(\mu sin\theta +cos\theta )$ is maximum when its derivative is zero.
$\left(\frac{d}{d\theta }\right)(\mu sin\theta +cos\theta )=0$

$\mu cos\theta -sin\theta =0$

$\theta =ta{n}^{-1}\mu$

So, $P=\frac{\mu mg}{\mu Sin\theta +cos\theta }$

$=\frac{\mu mglcos\theta }{\frac{(\mu sin\theta +cos\theta )}{cos\theta }}$

By dividing the numerator and denominator by $cos\theta$, we get
$P=\frac{\mu mgsec\theta }{1+\mu tan\theta }$

$=\frac{\mu mgsec\theta }{1+ta{n}^2\theta }$

$=\frac{\mu mg}{sec\theta }$

$=\mu mg/\sqrt{1+ta{n}^2\theta }$

$=\frac{\mu mg}{(1+{\mu }^2)}$

Hence minimum force is $\frac{\mu mg}{\sqrt{1+u^2}}$ at an angle $q=ta{n}^{-1}\mu$.