Question

A block of mass M is placed on surface with coefficient of friction $\mu$. M is connectedby a string which passes over an ideal pulley and other end of string is connected to a point mass $m=\frac{M}{8}$. Length of string between point mass m and pulley is l. Point mass is released at an angle of $\theta ={60}^{\circ }$ with vertical as shown in figure then.

• A. Minimum value of $\mu$ so that M does not slip is 0.25
• B. Maximum tension in string is2 mg
• C. If $\mu =0.1$ net work done by tension is not zero
• D. Minimum tension in string is $\frac{mg}{2}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A Minimum value of $\mu$ so that M does not slip is 0.25
B Maximum tension in string is2 mg
D Minimum tension in string is $\frac{mg}{2}$

$f_l=\mu Mg\dots \dots \left(1\right)$

By COE, when the mass reaches bottom, its velocity is $V=\sqrt{gl}$

Tension at the bottom is,
$T-mg=\frac{m{V}^2}{l}\Rightarrow T=2mg$
Tension is maximum at bottom as speed is maximum at bottom.
$f_l=T_{max}\Rightarrow \mu Mg=2mg$

$\mu =\frac{2m}{M}=0.25$
Minimum tenion happens when $\theta ={60}^{\circ }$
$\Rightarrow T_{min}=mg\cos {60}^{\circ }=\frac{mg}{2}$