Question

A block of mass $m$ is placed on the top of a $6 \, kg$ cart such that the time period of the system is $0.75 \, s$ assuming there is no slipping. If the cart is displaced by $50 \, mm$ from its equilibrium position and released, then the coefficient of static friction $\mu _{s}$ between block and cart is just sufficient to prevent the block from sliding. The value of $m$ and $\mu _{s}$ respectively are (Take $g=9.8 \, m/s^{2}$ )

• A. $1.63kg;0.251$
• B. $2.55kg;0.385$
• C. $3.42kg;0.632$
• D. $4.28kg;0.876$

Answer 1

Answer: (B)

$ \text{T} = 2 \pi \sqrt{\frac{\text{m} + 6}{\text{600}}} \, \, \, \left(\right. \text{T} = 2 \pi \sqrt{\frac{\text{m}}{\text{k}}} \left.\right)$
$\text{or} \text{0.75} = 2 \pi \sqrt{\frac{\text{m} + 6}{\text{600}}}$
$\therefore \quad \mathrm{m}=\frac{(0.75)^2 \times 600}{(2 \pi)^2}-6$
= 2.55 kg
Maximum acceleration of SHM is,
a_max = ω²A (A = amplitude)
i.e., maximum force on mass 'm' is m ω² A which is being provided by the force of friction between the mass and the cart. Therefore,
$\mu_{\mathrm{s}} \mathrm{mg} \geq \mathrm{m} \omega^2 \mathrm{~A}$
$
or $\mu_{ s } \geq \frac{\omega^{2} A }{ g }$
Or $\mu_{ s } \geq\left(\frac{2 \pi}{ T }\right)^{2} \cdot \frac{ A }{ g }$
Or $\mu_{ s } \geq\left(\frac{2 \pi}{0.75}\right)^{2}\left(\frac{0.05}{9.8}\right) \quad( A =50 mm )$
Or $\mu_{ s } \geq 0.358$
Thus, the minimum value of $\mu_{ s }$ should be 0.358