Suppose that in the same Atwood setup another string is attached to the bottom of m1...

Suppose that in the same Atwood setup another string is attached to the bottom of m₁ and a constant force f is applied, retarding the upward motion of m₁. If m₁ = 5.70 kg and m₂ = 11.40 kg, what value of f will reduce the acceleration of the system by 50%?

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Suppose that another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 6.75 kg and m2 = 11.25 kg, what value of f will reduce the acceleration of the system by 55%?

From Newtons' laws
m1a = T - m1g
and
m2a = m2g - T
Then
m1a = m2g - m2a - m1g
then acceleration
(m1 + m2 ) a = m2 g -m1g
a = (m2 - m1)g / m1 + m2
= (11.25 - 6.75) 9.8 / 11.25+6.75
= 2.45m/s^2
when the acceleration becomes a = 0.55a = 1.35m/s^2
m1a = T - m1g - f
and
m2a = m2g - T
T = m2g - m2a
then
m1a = m2g - m2a - m1g - f
(m1 + m2)a = (m2 - m1)g - f
therefore the applied force
f = (m2 - m1)g - (m1 + m2)a
= (11.25 - 6.75) (9.8) - ( 11.25 + 6.75) (1.35)
= 19.8 N

genius_generous answered 1 hour ago

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