Question

A block A of mass $5\text{kg}$ is placed on a table and connected to another block B of mass $3\text{kg}$ with a light string passing over a frictionless pulley. The coefficient of friction between the table the block A is $0.4$. The block B is suddenly dropped and the string breaks after $3$ seconds. What is the total distance covered by block A?

• A. $5.625\text{m}$
• B. $1.75\text{m}$
• C. $7.375\text{m}$
• D. $6.5\text{m}$

Answer 1

The correct option is C $7.375\text{m}$


From the free body diagrams, we have
$m_{B}g-T=m_{B}a...\left(i\right)$



Balancing force along $y$ axis.
$N=m_{A}g$
$f_k=\mu N=\mu m_{A}g$
Balancing force along $x$ axis.
$T-f_k=T-\mu m_{A}g=m_{A}a...\left(ii\right)$

Adding equation $\left(i\right)$ and $\left(ii\right)$
$g(m_B-\mu m_A)=a(m_A+m_B)$

$\Rightarrow a=\frac{g(m_B-\mu m_A)}{(m_A+m_B)}$

$=\frac{10(3-0.4\times 5)}{(3+5)}$

$=1.25{\text{m/s}}^2$



During the first 3 seconds, the block A travels a distance of
$s=ut+\frac{1}{2}a{t}^2$
$=0+\frac{1}{2}\times 1.25\times 3^2$
$=5.625\text{m}$


After the string gets cut, the block A will have an initial velocity
$u_1=0+1.25\times 3=3.75\text{m/s}$
and an acceleration $a_1=-\mu g=-0.4\times 10=-4{\text{m/s}}^2$

Here the negative sign indicates that the body will slow down after the string is cut due to friction.

The distance that the block covers after the string is cut,
$s_1=\frac{v^2-u_1^2}{2a_1}$

$=\frac{0-{3.75}^2}{2(-4)}$

$=1.75\text{m}$
Total distance$=(s+s_1)=(5.625+1.75)\text{m}=7.375\text{m}$

$\therefore \text{The block covers a total distance of}7.375\text{m}.$