Atwood's Machine Acceleration Rate and Maximum Value

[arenzob]

Homework Statement

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 1.3 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. At what rate is the acceleration magnitude of the containers changing at (a)t = 0 and (b)t = 3 s? (c) When does the acceleration reach its maximum value?

Homework Equations

T - m₁ g= m₁ a
T - m₂ g= -m₂ a
a= m₂ - m₁ g/ m₂ + m₁

3. Attempt to solve the problem:
I have been trying to solve this problem, but I can get the same answer as the book.
The answer in the the book is:
a)0.653m/s³
b)0.896m/s³
c)6.50s

a=[m2-m1/m2+m1]g
da/dt = -[(2.8-m1)/(2.8+m1)^2](dm1/dt)g
after this, i don't know what to do anymore. Am i doing it right?Thanks! :)

 

[gneill]

Show (at least) one of your attempts.

 

[gneill]

a=[m2-m1/m2+m1]g
da/dt = -[(2.8-m1)/(2.8+m1)^2](dm1/dt)g
after this, i don't know what to do anymore. Am i doing it right?

I don't think that the derivative of the acceleration is going to help you. Look instead to create an expression for the mass of M1 with respect to time to use in the acceleration equation.

 

[arenzob]

I have solved my problem.
All I did was that:
da/dt = da/dm1 ⋅ dm1/dt = [(m2+m1)(-1)-(m2-m1)(1)]/(m2+m1)² ⋅ dm1/dt
da/dt = g [-2m2/(m2+m1)²] ⋅ dm/dt
da/dt = (-3.92)[m2/(m2+m1)^2]
a) at t=0; m1= 1.3
da/dt= 0.653 m/s^3
b) at t=3; m1=1.3-0.2(3)=0.7
da/dt= 0.896 m/s^3
c) if m=0; a=max
m - 0.2t =0
t=1.3/0.2
t=6.5s