Question

A bead sliding on a wire has a speed of 210 cm/s at A. If friction forces are negligible, what will

be the speed at pt. B and C?

Answer 1

To be given in question

Bead velocity in point A $v_{A}=2.1meter/sec$

To be asked in question

Point B and pointC Speed find out

$v_{B}=?;v_{C}=?$

We know that energy of conservation

at point B speed

$\frac{1}{2}m{v}_f^2+\frac{1}{2}m{v}_i^2+mg(h_{f}-h_{i}) =0$

At point B hight zero

$h_{f}=0$

$\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+mg(0-h_{i}) =0$

$v^2_{f}-v^2_{i} =2gh_{i}$

$v_{f}^2=2gh_{i}+v^2_{i}$

$v_{f}=\sqrt {2gh_{i}+v^2_{i}}$

Put value $v_{i}=v_{A}=2.1meter/sec$

$v_{f}=\sqrt{2gh+4.41} meter/sec$

Point B velocity $v_{B}=v_{f}=[v_{f}]_{B}$

$[v_{f}]_{B}=\sqrt{2gh+4.42} meter/sec$

Pont c velocity

$\frac {1}{2}mv^2_{f}-\frac{1}{2}mv^2_{i} +mgh_{f}-mgh_{i}=0$

$v^2_{f}-v^2_{i} =2g(h_{i}-h_{f})$

$v^2_{f}=v^2_{i} +2g(h_{i}-h_{f})$

$v_{f}=\sqrt {v^2_{i} +2g(h_{i}-h_{f})}$

$v_{f}=v_{c}=[v_{f}]_{c}$

$[v_{f}]_{c}=\sqrt{4.41+2g(h_{i}-h_{f}) }meter/sec$