Question

A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in newton) P=10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is

Answer 1

$N=mg=200N$
$(f_s{)}_{max}={\mu }_{s}N$
$0.4\times 200=80N$
upto $8sec[P=10t]\to$ no motion
$\Rightarrow$ from 8 sec to 10 sec
$f_k={\mu }_{k}N$
$=0.2\times 200=40N$
Average applied force
$=\frac{P_{max}+P_{min}}{2}=\frac{100+80}{2}=90N$
Average resultant force = Avg. applied force - kinetic friction = 90 - 40 = 50 N
Average acceleration
$\frac{50\times 9.8}{200}$
$\Rightarrow V=u+at$
$u=0$
$a=\frac{50\times 9.8}{200}$
t = 2 second
$\therefore V=\frac{50\times 9.8}{200}\times 2=4.9\frac{m}{s}$