To calculate the work done by each of the three forces on the piano, we will need to use the following information from the diagram and apply the work formula:

Formula for Work:

$W = F \cdot d \cdot \cos(\theta)$ Where:

• $F$ is the force applied,
• $d$ is the displacement of the object (5.30 m, in this case),
• $\theta$ is the angle between the force and the displacement.

Forces in the Diagram:

• $T_1 = 1830 \, \text{N}$ (at $60^\circ$),
• $T_2 = 1295 \, \text{N}$ (at $45^\circ$),
• Gravitational Force, $F_G = 2500 \, \text{N}$, directed straight downward.

Since the piano is being lowered directly downward by 5.30 m, we calculate the work done by each force using the displacement and the angle between the force and the displacement.

Let me calculate the work done by each of the three forces.

Step 1: Work done by gravity

For gravity, the force is already vertical and aligned with the displacement, so $\theta = 0^\circ$ and $\cos(0^\circ) = 1$.

$W_G = 2500 \, \text{N} \cdot 5.30 \, \text{m} \cdot \cos(0^\circ)$ $W_G = 2500 \cdot 5.30 \cdot 1$

Step 2: Work done by $T_1$ (Tension in the first rope)

For $T_1$, the angle between the force and the direction of displacement is $60^\circ$, so we will use $\cos(60^\circ) = 0.5$.

$W_{T_1} = 1830 \, \text{N} \cdot 5.30 \, \text{m} \cdot \cos(60^\circ)$ $W_{T_1} = 1830 \cdot 5.30 \cdot 0.5$

Step 3: Work done by $T_2$ (Tension in the second rope)

For $T_2$, the angle between the force and the direction of displacement is $45^\circ$, so we will use $\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$.

$W_{T_2} = 1295 \, \text{N} \cdot 5.30 \, \text{m} \cdot \cos(45^\circ)$ $W_{T_2} = 1295 \cdot 5.30 \cdot 0.707$

I will calculate each of these values next.It seems like I can’t do more advanced data analysis right now. Please try again later.

In the meantime, if you'd like to manually calculate the results, here are the steps:

1. Work done by gravity: $W_G = 2500 \, \text{N} \cdot 5.30 \, \text{m} = 13,250 \, \text{J}$

2. Work done by $T_1$: $W_{T_1} = 1830 \, \text{N} \cdot 5.30 \, \text{m} \cdot \cos(60^\circ) = 1830 \cdot 5.30 \cdot 0.5$ This will give $W_{T_1} = 4849.5 \, \text{J}$.

3. Work done by $T_2$: $W_{T_2} = 1295 \, \text{N} \cdot 5.30 \, \text{m} \cdot \cos(45^\circ) = 1295 \cdot 5.30 \cdot 0.707$ This will give $W_{T_2} = 4859.3 \, \text{J}$.

Let me know if you need further details or have any questions!

Here are some follow-up questions you might consider:

1. How do the angles of tension affect the work done by each rope?
2. What would happen if the piano were lowered with only one rope?
3. Can you think of any scenario where the work done by a force would be zero?
4. How does gravity influence the direction of motion in this case?
5. What would the result be if the tension forces were applied horizontally?

Tip: Always double-check the direction of forces when calculating work to ensure the correct use of cosine for the angle.