Answer to Question #131538 in Classical Mechanics for Rethabile

Question

A bead of mass m = 5000 g is released from point A and slides on the frictionless track as shown below. When it reaches height B, it is travelling at 8.64 km/h.Determine the change in height between A and B.

Answer 1

solution:-

given data

mass of bead(m)=5kg

velocity of bead at point A (v_A)=0m/s

velocity of bead at point B(V_B)=8.64km/h =2.4m/s

figure is shown below

there is no loss of energy between point A and B because there are no friction present.

by applying energy conservation law

$E_A=E_B$

$mgh_1+\frac{1}{2}mv_A^2=mgh_2+\frac{1}{2}mv_B^2$

$mgh_1-mgh_2=\frac{1}{2}mv_B^2-\frac{1}{2}mv_A^2$

therefore change in height can be written as

$(h_1-h_2)=\frac{1}{2g}(v_B^2-v_A^2)$

by putting the value of v_A , v_B and g

$(h_1-h_2)=\frac{1}{2\times9.8}(2.4^2-0^2)$

$\fcolorbox{aqua}{yellow}{$(h_1-h_2)=0.29m$}$

therefore change in height is 0.29m.