Atwood machine Force & Motion problem

[gomerpyle]

Homework Statement

Figure 5-53 shows Atwood's machine, in which 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 1.30kg and container 2 has mass 2.80kg, but container 1 is losing mass at the constant rate of 0.200kg/s. At what rate is the acceleration magnitude of the containers changing at (a) t = 0 and (b) t = 3.00s? (c) When does the acceleration reach its maximum value?

Homework Equations

I know this has something to do with F=ma but I'm really lost on this one.

The Attempt at a Solution

Don't know where to start, really.

 

[anathema]

Hello there,

First, let's define the variables in this problem:

m1 = mass of container 1 (initially 1.30kg)
m2 = mass of container 2 (initially 2.80kg)
a = acceleration of the system
t = time
dm1/dt = rate of change of mass of container 1 (given as -0.200kg/s)

Now, let's apply Newton's second law (F=ma) to the system:

For container 1:
F1 = m1*a
As the container is losing mass, the force acting on it will also decrease over time. So, the force acting on container 1 at time t will be:
F1 = (m1 - dm1/dt*t)*a

For container 2:
F2 = m2*a

Now, let's consider the forces acting on the system at time t = 0:
F1 = (1.30 - 0.200*0)*a = 1.30a
F2 = 2.80*a

Since the two containers are connected by a cord, the tension in the cord will be equal to the force acting on container 2 (F2). So, we can equate the two forces:
F1 = F2
1.30a = 2.80*a
a = 2.80/1.30 = 2.15 m/s^2

So, at t = 0, the acceleration of the system is 2.15 m/s^2.

Now, let's consider the forces acting on the system at time t = 3.00s:
F1 = (1.30 - 0.200*3)*a = 0.70a
F2 = 2.80*a

Again, equating the two forces:
F1 = F2
0.70a = 2.80*a
a = 2.80/0.70 = 4.00 m/s^2

So, at t = 3.00s, the acceleration of the system is 4.00 m/s^2.

To find the maximum acceleration, we need to find the point where the acceleration is changing from positive to negative (i.e. the point where the acceleration is at its peak and starts to decrease). This can be found by setting the rate of change of acceleration (

 

[m2003]

I would approach this problem by first identifying the key variables and equations involved. In this case, the key variables are the masses of the containers and the acceleration of the system, and the key equation is F=ma.

To begin, we can calculate the initial acceleration of the system at t=0 by using the equation F=ma, where F is the net force acting on the system and m is the total mass of the system. In this case, the net force is the difference between the weight of container 1 (1.30kg) and the weight of container 2 (2.80kg), since the tension in the cord is equal to the weight of the containers (neglecting the mass of the cord and pulley). This gives us an initial acceleration of 1.30kg - 2.80kg = -1.50kg.

To answer part (a), we need to find the rate of change of acceleration at t=0. This can be done by differentiating the initial acceleration equation with respect to time. Since the mass of container 1 is decreasing at a rate of 0.200kg/s, this affects the net force on the system, and therefore the acceleration. The new net force is given by the difference between the weight of container 1 (1.30kg - 0.200kg/s*t) and the weight of container 2 (2.80kg), resulting in a new acceleration equation of (1.30kg - 0.200kg/s*t) - 2.80kg = -1.50kg + 0.200kg/s. Differentiating this equation with respect to time gives us the rate of change of acceleration at t=0, which is 0.200kg/s.

For part (b), we can use the same approach as in part (a), but this time we are looking for the rate of change of acceleration at t=3.00s. Using the same acceleration equation as in part (a), but substituting t=3.00s, we get a new acceleration of -1.50kg + 0.200kg/s*3.00s = -0.90kg. Differentiating this equation with respect to time gives us a rate of change of acceleration at t=3.00s of 0.200kg/s.

Finally, to find the time at which the acceleration reaches its maximum value