Question

Figure 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 1.30 kgand container 2 has mass 2.80 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 = 0and (b) At what rate is the acceleration magnitude of the containers changing at t = 3.00s?(c) When does the acceleration reach its maximum value?

Short answer

(a) The rate of change of acceleration of the containers at t = 0s is$0.653m/s^3$

(b) The rate of change of acceleration of the containers at t = 3s is$0.896m/s^3$

(c) At t = 6.5s the acceleration reaches its maximum value.

Step-by-step solution

Given information

It is given that,

The mass of the ice block is$M_1=1.3kg$

The mass of the ice block is$M_2=2.8kg$

The rate of change of mass,$\frac{d{M}_1}{dt}=0.2\frac{kg}{s}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Using Newton’s second law, the acceleration and from that,the rate of change of acceleration of the containers can be found.

Formulae:

Newton’s second law is,

$F_{net}=\sum _{}Ma$ (i)

$\frac{da}{dt}=\frac{da}{dM}\times \frac{dM}{dt}$ (ii)

where,$F_{net}$ is the net force, Mis mass and a is an acceleration.

(a) Determining the rate of change of acceleration of the containers at t = 0s

Apply Newton’s second law,

$T-M_{1}g=M_{1}a$ (iii)

$M_{2}g-T=M_{2}a$ (iv)

Solving equations (iii) and (iv),

$a=\left(\frac{M_2-M_1}{M_1+M_2}\right)g$

At t = 0 ,$M_1=1.3kg$ .The rate of change of mass,

$\frac{d{M}_1}{dt}=0.2\frac{kg}{s}$

So,its rate of change of acceleration is,

$\frac{da}{dt}=\frac{da}{d{M}_{10}}\times \frac{d{M}_1}{dt}$

Using division rule and considering$M_2$as constant,

$$\begin{gathered} \frac{da}{dt}=\left(\frac{\left\{\left(-1\right)\left(M_1+M_2\right)-\left(M_2-M_1\right)\right\}}{\left(M_1+M_2\right)}\right)\frac{d{M}_1}{dt} \\ \frac{da}{dt}=-\frac{2M_{2}g}{\left(M_1+M_2\right)}\times \frac{d{M}_1}{dt} \\ \frac{da}{dt}=\frac{-2\left(2.8\right)\left(9.8\right)\left(-0.2\right)}{{\left(2.8+1.3\right)}^2} \\ \frac{da}{dt}=0.653m/s^3 \end{gathered}$$

Therefore, the rate of change of acceleration of the containers at t = 0 is$0.653m/s^3.$

(b) Determining the rate of change of acceleration of the containers at  t = 3s

At t = 3 s

$$\begin{gathered} M_1=M_1+\left(\frac{d{M}_1}{dt}\right)t \\ {M}_1=1.3+\left(-2\right)\left(3\right) \\ =0.7kg \end{gathered}$$

The rate of change of mass,

$\frac{d{M}_1}{dt}=0.2kg/s$

So, its rate of change of acceleration is,

$$\begin{gathered} \frac{da}{dy}=\frac{da}{d{M}_1}+\frac{d{M}_1}{dt} \\ \frac{da}{dt}=\left[\frac{\left\{\left(-1\right)\left(M_1+M_2\right)-\left(M_2-M_1\right)\right\}}{{\left(M_1+M_2\right)}^2}\right]\frac{d{M}_1}{dt} \\ \frac{da}{dt}=-\frac{2M_{2}g}{{\left(M_1+M_2\right)}^2}\times \frac{d{M}_1}{dt} \\ \frac{da}{dt}=\frac{-2\left(2.8\right)\left(9.8\right)\left(-0.2\right)}{{\left(2.8+0.7\right)}^2} \\ \frac{da}{dt}=0.896m/s^2 \end{gathered}$$

Hence, the rate of change of acceleration of the containers at t = 3s is $0.896m/s^2.$

(c) Determining when the acceleration reaches its maximum

The acceleration reaches its maximum when$M_1=0$ , so,

$$\begin{gathered} M_1+\left(\frac{d{M}_1}{dt}\right)t=0 \\ 1.3-\left(0.2\right)t=0 \\ t=6.5s \end{gathered}$$

Hence, at$t=6.5s$ the acceleration reaches its maximum value.