Question

A block is projected with speed $u$ on the rough surface of coefficient of friction $\mu$ stopping distance is s. If coefficient of friction becomes $\frac{\mu }{4}$ and projected speed is same . The stopping distance becomes

• A. $4$s
• B. s
• C. $\frac{s}{4}$
• D. $2$s

Answer 1

Answer: (A)

$4$s

From $A$ to $B$$V{v}^2=u^2+2as$

$0=u^2+2(-\mu g)s$

$⟹s=\frac{u^2}{2\mu g}$

stopping distance$\propto$ $\frac{1}{\mu }$

$\frac{s_1}{s_2}=\frac{{\mu }_1}{{\mu }_2}$

$⟹\frac{s}{s_{req}}=\frac{1}{4}$

$⟹s_{req}=4s$