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A system is released as shown. Neglect drag, friction, and the mass of the string and pulley.

Calculate the unknown(s) in each of the following cases.

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
M \, \text{[kg]} & m \, \text{[kg]} & a \, \left[\text{m/s}^2\right] \\
\hline
13 & 6 & 11.4 \\
\hline
13 & 9 & 1.4 \\
\hline
13 & 12 & \\
\hline
16 & 12 & \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnswer
In this problem, we are given a table that provides the masses [tex]\( M \)[/tex] and [tex]\( m \)[/tex] of two objects for several cases, along with their resulting accelerations [tex]\( a \)[/tex]. We need to calculate the unknown accelerations for the last two cases. The given information and the required unknowns can similarly be analyzed using Newton's second law in the context of two-body systems connected by a string over a frictionless pulley.

The formula to calculate the acceleration [tex]\( a \)[/tex] of the system is:
[tex]\[ a = \frac{(M - m) \cdot g}{(M + m)} \][/tex]

where:
- [tex]\( g \)[/tex] is the acceleration due to gravity, approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
- [tex]\( M \)[/tex] and [tex]\( m \)[/tex] are the masses in [tex]\( \text{kg} \)[/tex].

We are given:
1. [tex]\( M = 13 \, \text{kg}, m = 6 \, \text{kg}, a = 11.4 \, \text{m/s}^2 \)[/tex]
2. [tex]\( M = 13 \, \text{kg}, m = 9 \, \text{kg}, a = 1.4 \, \text{m/s}^2 \)[/tex]
3. [tex]\( M = 13 \, \text{kg}, m = 12 \, \text{kg}, a = \text{?} \)[/tex]
4. [tex]\( M = 16 \, \text{kg}, m = 12 \, \text{kg}, a = \text{?} \)[/tex]

Let's solve for the unknown accelerations [tex]\( a \)[/tex] in the last two cases.

Case 3:
[tex]\[ M = 13 \, \text{kg}, m = 12 \, \text{kg} \][/tex]

Using the formula:
[tex]\[ a_3 = \frac{(13 - 12) \cdot 9.8}{(13 + 12)} \][/tex]

Breaking it down step by step:
- Numerator: [tex]\( 13 - 12 = 1 \)[/tex]
- Denominator: [tex]\( 13 + 12 = 25 \)[/tex]

So,
[tex]\[ a_3 = \frac{1 \cdot 9.8}{25} = \frac{9.8}{25} \approx 0.392 \, \text{m/s}^2 \][/tex]

Case 4:
[tex]\[ M = 16 \, \text{kg}, m = 12 \, \text{kg} \][/tex]

Again, using the formula:
[tex]\[ a_4 = \frac{(16 - 12) \cdot 9.8}{(16 + 12)} \][/tex]

Step by step:
- Numerator: [tex]\( 16 - 12 = 4 \)[/tex]
- Denominator: [tex]\( 16 + 12 = 28 \)[/tex]

So,
[tex]\[ a_4 = \frac{4 \cdot 9.8}{28} = \frac{39.2}{28} \approx 1.400 \, \text{m/s}^2 \][/tex]

Therefore, the unknown accelerations for the given masses are:
- For [tex]\( M = 13 \, \text{kg} \)[/tex] and [tex]\( m = 12 \, \text{kg} \)[/tex]: [tex]\( a_3 \approx 0.392 \, \text{m/s}^2 \)[/tex]
- For [tex]\( M = 16 \, \text{kg} \)[/tex] and [tex]\( m = 12 \, \text{kg} \)[/tex]: [tex]\( a_4 \approx 1.400 \, \text{m/s}^2 \)[/tex]
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