A block of mass 2 kg placed on a rough horizontal surface having coefficient of friction 0.25 is acted upon by a force of
10 N as shown below. The distance travelled by the block in the first five seconds is

The correct option is B 31.25 m

Mass, $m=2kg$
External force, $F=10N$
Normal reaction, $N=20N$
Frictional force, $f=\mu R$
Here, $\mu$ = coefficient of friction = 0.25
$\Rightarrow R=mg=2\times 10=20N$
$\Rightarrow f=\mu R=0.25\times 20=5N$
Net force acting on the block, $F_{net}=F–f$
= 10 N – 5N
= 5 N
According to second law of motion:
$\Rightarrow F_{net}=ma=5N$
$\Rightarrow a=\frac{5}{m}=\frac{5}{2}=2.5m/s^2$

Since initially the block is not moving, its initial velocity, u is '0'.
Distance travelled by the block in 5sec,
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=0\times t+\frac{1}{2}\left(\frac{5}{2}\right)(5{)}^2$
$\Rightarrow s=\frac{125}{4}=31.25m$
The displacement in the first five seconds is 31.25 m.
Hence, the correct answer is option (2).