Question

A small, spherical bead of mass 2.50 g is released from rest at t = 0 from a point under the surface of a viscous liquid. The terminal speed is observed to be vT = 1.98 cm/s.

(a) Find the value of the constant b in the equation R with arrow = −bv with arrow.

______________________________________N·s/m

(b) Find the time t at which the bead reaches 0.632vT.

__________________s

(c) Find the value of the resistive force when the bead reaches terminal speed.

__________________________N

Answer 1

(. t = 0.00202 57 t -os er 14* 2.02 ms Sol:- given m=2.54-0.0025 Terminal spoed (V1)= 1.98 emls Vy = 0.0178 m/s a) The forces acting on bead are - the weight acting downward - and air drag (R) acts upward a. The net force will be - Efy = mg-bNama → but when the bead reaches terminal Velocity its acceleration a=0. @ At terminal dring force is equal to weight of the load so u R = mg = 0.0025×9.8 [R = 0.0245 N Hopt Oill get VOTE mg-bq=o > b= mg/up = 0.0025 x 7,8 0.0198 b = 1.237 N. Ibn 1.24 Nism b) given = 0.632 VT le we have lo. V-VII- => 0.632V = V1[1-e'tlo] => e-+T 1-0,632 = 0.368 > - terz' In (o.368) -bt - In (0.368) |--TEM). > t=-min (0.368) b = -0.0025 0.9996) 1-237 =0.00202 s dest