Question

# A block of mass 2 kg is on a horizontal surface. The coefficient of static and kinetic friction are 0.6 and 0.2. The minimum horizontal force required to start the motion is applied and if it is continued, the velocity acquired by the body at the end of the 2nd second is?

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Solution

## Step1: Given dataThe mass of the block, $m=2kg$The coefficient of static and kinetic friction are ${\mu }_{s}=0.6,{\mu }_{k}=0.2$Step2: Formula usedThe first equation of motion says,$v=u+at\left[wherev=finalvelocity,u=initialvelocity,a=acceleration,t=time\right]$Step3: Calculating net force on the bodyTo start the body, one has to overcome the threshold of static friction ${F}_{s}={\mu }_{s}mg=0.6×2×10=12N\left[As,g=10\right]$Now once the body starts moving the force required to overcome kinetic friction ${F}_{K}={\mu }_{k}mg=0.2×2×20=4N$The net force on the body, $F={F}_{s}-{F}_{k}=12-4=8N$Step4: Calculating the velocity after 2 sec$a=\frac{f}{m}\left[a=acceleration,f=force,m=mass\right]\phantom{\rule{0ex}{0ex}}a=\frac{8}{2}=4m/{s}^{2}$We need to start the body so the initial velocity will be $u=0$$v=u+at\phantom{\rule{0ex}{0ex}}v=0+4×2\phantom{\rule{0ex}{0ex}}v=8m/s$Hence, the minimum force required to start the body is 12 N, and the velocity after 2 sec will be 8 m/s.

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