6. 0.43/1.25 polnts | Previous Answers SerPSET9 5.P050. No In the Atwood machine

Question

6. 0.43/1.25 polnts | Previous Answers SerPSET9 5.P050. No In the Atwood machine shown below, m1 -2.00 kg and m2 6.30 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at v2.35 m/s downward r1 (a) How far will m descend below its initial level? 2.20 Your response differs from the correct answer by more than 100%, m ) Find the veloity of 1 m/s magnitude direction upward

Explanation / Answer

given

m1 = 2 kg

m2 = 6.3 kg

given frictionless pulley

let acceleration be mass m2 be a acting downwards

hence

from force balance

let tension in the stirng be T

then

m2*g - T = m2*a

T - m1*g = m1*a

(m2 - m1)g = (m1 + m2)a

a = 0.51 m/s/s

intiail speed given to the lighter object, u = -2.35 m/s

hence

a.

when this object comes to rest, it has decended distance d

hence

2*a*d = u^2

d = 5.3298546 m

b. velcoity of m1 after 1.8 s be v

v = u + at

v = -2.35 + a*t

v = -1.432 m/s ( in downwards direction)

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