Question

The two ropes are used to lower a 255 kg piano 5.00 m from a second-story window to the ground. How much work is done by each of the three forces?

Short answer

$\overrightarrow{{\mathrm{F}}_{\mathrm{G}}}=2.00\times {10}^{4}J$

$\overrightarrow{{\mathrm{T}}_1}=-1.27\times {10}^{4}J$

$\overrightarrow{T_2}=-7.326\times {10}^{4}J$

Step-by-step solution

Step 1. Given Information

We have, $\mathrm{W}=\mathrm{Fdcos}\left(\mathrm{\theta }\right)$

The work done by all the ropes by

$$\begin{gathered} \overrightarrow{{\mathrm{F}}_{\mathrm{G}}}=2500\times 8\times \cos \left(0°\right) \\ \overrightarrow{{\mathrm{F}}_{\mathrm{G}}}=20000J \\ \overrightarrow{{\mathit{F}}_{\mathit{G}}}=2.00\times {10}^{\mathit{4}}J \end{gathered}$$

For $\overrightarrow{{\mathrm{T}}_1}$,

$$\begin{gathered} \overrightarrow{{\mathrm{T}}_1}=1830\times 8\times \cos \left(150°\right) \\ \overrightarrow{{\mathrm{T}}_1}=14640\times -0.866 \\ \overrightarrow{{\mathrm{T}}_1}=-1.27\times {10}^4\mathrm{J} \end{gathered}$$

And for $\overrightarrow{{\mathrm{T}}_2}$,

$$\begin{gathered} \overrightarrow{{\mathrm{T}}_2}=1295\times 8\times \cos \left(135°\right) \\ \overrightarrow{{\mathrm{T}}_1}=10360\times -0.99 \\ \overrightarrow{{\mathrm{T}}_1}=-7.326\times {10}^4\mathrm{J} \end{gathered}$$