Question

A small block of mass m is released from rest from point D and slides down DGF and reaches the point F with speed $v_F$. The coefficient of kinetic friction between block and both the surfaces DG and GF is $\mu$, the velocity $v_F$ is:

• A. $\sqrt{2g(y-x)}$
• B. $\sqrt{2g(y-\mu x)}$
• C. $\sqrt{2gy}$
• D. $\sqrt{2g(y^2+x^2)}$

Answer 1

Answer: (B)

$\sqrt{2g(y-\mu x)}$

Here $mgy-\frac{1}{2}m{v}_F^2=f_2{s}_1+f_2{s}_2$

i.e. $\frac{1}{2}m{v}_V^2=mgy-\mu mg\cos \beta S_1-\mu mg\cos \alpha S_2$

or $\frac{1}{2}{v}_F^2=g(y-\frac{\mu x_1{S}_1}{S_1}-\frac{\mu x_2{S}_2}{S_2})$

or $v_F=\sqrt{2g(y-\mu x)}$