Answered

The force exerted by a pulley is the sum of the tension and the force of gravity. Since a pulley is used to lift something, the force is negative. The tables show the tension, [tex]\(T(m)\)[/tex], and the force of gravity, [tex]\(G(m)\)[/tex], for objects of various masses, [tex]\(m\)[/tex] in grams, all accelerating upward at [tex]\(2.5 \, m/s^2\)[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$T(m)$ (Newtons) & 1.25 & 3 & 6.5 & 8.5 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$G(m)$ (Newtons) & 4.9 & 11.76 & 25.48 & 33.32 \\
\hline
\end{tabular}
\][/tex]

The force of the rope is equal to the negative of the sum of the rope's tension and the force of gravity.

Which table shows the values when combining the functions to find the force of the rope?

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$-(T(m)+G(m))$ (Newtons) & 6.15 & 14.76 & 31.98 & 41.82 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$-(T(m)+G(m))$ (Newtons) & -6.15 & -14.76 & -31.98 & -41.82 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$-(T(m)+G(m))$ (Newtons) & -3.65 & -8.76 & -18.98 & -24.82 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$m$ (grams) & 0.5 & 1.2 & 2.6 & 3.4 \\
\hline
$-(T(m)+G(m))$ (Newtons) & 3.65 & 8.76 & 18.98 & 24.8 \\
\hline
\end{tabular}
\][/tex]



Answer :

GinnyAnswer
To find the force exerted by the pulley on the objects, we start with the given values for tension ([tex]\( T(m) \)[/tex]) and the force of gravity ([tex]\( G(m) \)[/tex]). We need to calculate the negative sum of these values for various masses ([tex]\( m \)[/tex]). The force of the rope is defined as:

[tex]\[ \text{Force of the rope} = - (T(m) + G(m)) \][/tex]

Let's apply this formula step-by-step for each mass:

1. For [tex]\( m = 0.5 \)[/tex]:
[tex]\[ T(0.5) = 1.25 \][/tex]
[tex]\[ G(0.5) = 4.9 \][/tex]
[tex]\[ \text{Force of the rope} = - (1.25 + 4.9) = - (6.15) = -6.15 \][/tex]

2. For [tex]\( m = 1.2 \)[/tex]:
[tex]\[ T(1.2) = 3 \][/tex]
[tex]\[ G(1.2) = 11.76 \][/tex]
[tex]\[ \text{Force of the rope} = - (3 + 11.76) = - (14.76) = -14.76 \][/tex]

3. For [tex]\( m = 2.6 \)[/tex]:
[tex]\[ T(2.6) = 6.5 \][/tex]
[tex]\[ G(2.6) = 25.48 \][/tex]
[tex]\[ \text{Force of the rope} = - (6.5 + 25.48) = - (31.98) = -31.98 \][/tex]

4. For [tex]\( m = 3.4 \)[/tex]:
[tex]\[ T(3.4) = 8.5 \][/tex]
[tex]\[ G(3.4) = 33.32 \][/tex]
[tex]\[ \text{Force of the rope} = - (8.5 + 33.32) = - (41.82) = -41.82 \][/tex]

Combining these results into a table, we get:

[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $m$ & 0.5 & 1.2 & 2.6 & 3.4 \\ \hline $-(T(m)+G(m))$ & -6.15 & -14.76 & -31.98 & -41.82 \\ \hline \end{tabular} \][/tex]

Therefore, the table that correctly displays the combined functions to find the force of the rope is the second table:

[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $m$ & 0.5 & 1.2 & 2.6 & 3.4 \\ \hline $-(T(m)+G(m))$ & -6.15 & -14.76 & -31.98 & -41.82 \\ \hline \end{tabular} \][/tex]

Other Questions