Chapter 5: Q54P (page 120)

Figure 5-49 shows four penguins that are being playfully pulled along very slippery (frictionless) ice by a curator. The masses of three penguins and the tension in two of the cords are $m_{1} = 12 kg$ , $m_{3} = 15 kg$ , $m_{4} = 20 kg$ ,role="math" localid="1661176245206" $T_{2} = 111 N$ , and $T_{4} = 222 N$ . Find the penguin mass $m_{2}$ that is not given.

Short Answer

Expert verified

The mass of the penguin is $23 kg$ .

Step by step solution

Given data:

The mass of the first penguin, $m_{1} = 12 kg$

The mass of the third penguin, $m_{3} = 15 kg$

The mass of the fourth penguin, $m_{4} = 20 kg$

The tension in the cord, $T_{2} = 111 N$ , and

The tension in the cord, $T_{4} = 222 N$

Understanding the concept:

According to Newton's second law, an object's force and acceleration are both directly inversely proportional. The vector sum of each force exerted on the item is what is known as the net force.

Apply Newton’s second law to solve the problem.

The mass penguin m2 :

First, consider all the penguins (1 through 4, counting left to right) as one system, to which you can apply Newton’s second law:

$T_{4} = (m_{1} + m_{2} + m_{3} + m_{4}) a$

$222 N = (12 kg + m_{2} + 15 kg + 20 kg) \cdot a$ ….. (1)

Second, consider penguins 3 and 4 as one system, for which you have

$\begin{matrix} T_{4} - T_{2} = (m_{3} + m_{4}) a \\ 111 N = (15 kg + 20 kg) a \\ a = \frac{111 N}{35 kg} \\ a = 3.2 m / s^{2} \end{matrix}$

Substituting the above value into equation (1), you obtain,

$\begin{matrix} 222 N = (12 kg + m_{2} + 15 kg + 20 kg) \times 3.2 m / s^{2} \\ \frac{222 N}{3.2 m / s^{2}} = (47 kg + m_{2}) \\ m_{2} = 69.4 kg - 47 kg \\ m_{2} \approx 23 kg \end{matrix}$