Two red blood cells each have a mass of 9.05 × 10-14 kg and carry a negative cha

Question

Two red blood cells each have a mass of 9.05 × 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. One cell carries -3.00 pC and the other -3.50 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial speed would each need so that they get close enough to just barely touch? Assume that there is no viscous drag from any of the surrounding liquid.

__ m/s

What is the maximum acceleration of the cells as they move toward each other and just barely touch?

__ m/s2

Explanation / Answer

When cells are far away,

PE = 0

KE = 2 ( m v^2 /2 ) = m v^2

when cells are touching each other,

PE = kq1q2/ d

and d = 2r

PEf = (9 x 10^9 x 3 x 10^-12 x 3.50 x 10^-19) / (2 x 3.75 x 10^-6)

PEf = 1.26 x 10^-8 J

KE= 0

applying energy conservation,

m v^2 + 0 = 0 + 1.26 x 10^-8

(9.05 x 10^-14) v^2 = 1.26 x 10^-8

v = 373.13 m/s ................Ans

when distance is shortest.

d = 2r

Fmax = (9 x 10^9 x 3 x 10^-12 x 3.5 x 10^-12) / (3.75 x 10^-6)^2 = 6.72 x 10^-3 N

a_max = Fmax / m = (6.72 x 10^-3) / (9.05 x 10^-14) = 7.425 x 10^10 m/s^2

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