Question

$M$ is a fixed wedge. Masses $m_1$ and $m_2$ are connected by a light string. The wedge is smooth and the pulley is smooth and fixed. $m_1=10kg$ and $m_2=7.5kg$. When $m_2$ is just released, the distance it will travel in $2$ seconds is

• A. $7.5m$
• B. $2.8m$
• C. $6.0m$
• D. $4.0m$

Answer 1

The correct option is B $2.8m$

Draw $FBD$ of both blocks
Given: $m_1=10kg,m_2=7.5kg,t=2s$

Both blocks are attached to same string, hence both will have same magnitude of acceleration

Find acceleration of block

FORMULA USED: $a=\frac{\text{Net Force}}{\text{Total mass}}$

$a=\frac{m_{2}g-m_{1}g\sin {30}^{\circ }}{m1+m2}$

$a=\frac{75-\frac{100}{02}}{10+7.5}$
$=\frac{25}{17.5}=\frac{10}{7}m/s^2$

Find distance travelled by block

FORMULA USED: $s=\frac{1}{2}a{t}^2$
$s=\frac{1}{2}a{t}^2$

$=\frac{1}{2}\left(\frac{10}{7}\right){\left(2\right)}^2$

$=\frac{20}{7}\approx 2.8m$

Final anwer: (a)