Question

A block of mass \(m = (2 ~\text{kg})\) is placed on a rough horizontal surface and is being acted upon by a time dependant force \(F (N) = 2t\) (where \(t\) is in second). The coefficient of static friction between the block and the horizontal surface is \(\mu_s = 0.20\). The frictional force \(f\) developed between the block and the surface versus force \(F (N)\) plot is as shown.
The velocity of the block at \(t=4~\text{s}\) will be
\( 7 / l l l l l l l l l l l l l l l l \)

• A. $2.5\text{m/s}$
• B. $1\text{m/s}$
• C. $5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is D $3\text{m/s}$

Maximum friction force
$f_{\text{max}}={\mu }_{s}mg$
$=\left(0.2\right)\left(2\right)\left(10\right)=4\text{N}$
At $t=2\text{s},F\left(N\right)=4\text{N}$

Upto $t=2\text{s}$, block is in rest. After that relative motion starts between block and surface and kinetic friction acts between surfaces. Now from figure, force due to friction is $F_r=4-1=3$
Acceleration after two second is $a=\frac{F}{m}$
$a=\frac{2t-3}{2}$
$\frac{dv}{dt}=\frac{2t-3}{2}$
${\int }_0^{v}dv={\int }_2^4\left(\frac{2t-3}{2}\right)dt$
$v=3\text{m/s}$