- #1
theunloved
- 43
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two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.8 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.200 kg/s.
a) At what rate is the acceleration magnitude of the containers changing at t = 0?
(b) At what rate is the acceleration magnitude of the containers changing at t = 3.00 s?
(c) When does the acceleration reach its maximum value?[PLAIN]http://img5.imageshack.us/img5/50/q55u.jpg
Take upwards as positive
Second Newton's law
m1 : -m1 g + T = m1 a (1)
m2: -m2 g + t = -m2 a (2)
(1) - (2)--------> g(m2 - m1) = a (m1 + m2)
a = g(m2 - m1) / (m1 + m2)a. t = 0
m1 = 1.3kg
m2 = 2.8 kg
a = 3.58 m/s^2
b. t = 3s
m1 = 1.3 - 3 * 0.2 = 0.7 kg
m2 = 2.8 kg
a = 5.88 m/s^2
c. Ok, I don't know how to do this part. I believe that in order to get max. acceleration, m1 must be = 0, so a will be equal to g ? is that correct ?
a) At what rate is the acceleration magnitude of the containers changing at t = 0?
(b) At what rate is the acceleration magnitude of the containers changing at t = 3.00 s?
(c) When does the acceleration reach its maximum value?[PLAIN]http://img5.imageshack.us/img5/50/q55u.jpg
Take upwards as positive
Second Newton's law
m1 : -m1 g + T = m1 a (1)
m2: -m2 g + t = -m2 a (2)
(1) - (2)--------> g(m2 - m1) = a (m1 + m2)
a = g(m2 - m1) / (m1 + m2)a. t = 0
m1 = 1.3kg
m2 = 2.8 kg
a = 3.58 m/s^2
b. t = 3s
m1 = 1.3 - 3 * 0.2 = 0.7 kg
m2 = 2.8 kg
a = 5.88 m/s^2
c. Ok, I don't know how to do this part. I believe that in order to get max. acceleration, m1 must be = 0, so a will be equal to g ? is that correct ?
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