Question

b. if all of the energy from the student’s lunch did something useful, like lifting pianos weighing 5000. newtons to the top of a 10 - meter tall apartment building, how many pianos could be lifted with the energy from lunch? (ignore the energy radiated by the student.) complete the energy bar graph below to aid your solution.
energy conservation equation:

Explanation:

To solve this, we first assume the energy from the lunch (let's say from a typical lunch, but since it's not given, maybe from a common problem context, let's assume the energy from lunch is, for example, \( E = 2.5\times10^{6}\, \text{J} \) (common lunch energy, but wait, actually, maybe we missed the energy value? Wait, no, maybe in the original problem (not shown here), the energy from lunch is, say, \( 2.5\times10^{6}\, \text{J} \). Wait, no, the user's problem might have a typo or missing data? Wait, no, maybe in the original problem (the image), the energy from lunch is, for example, let's check the work done per piano.

Work done to lift one piano: \( W = F \times d \), where \( F = 5000\, \text{N} \), \( d = 10\, \text{m} \). So \( W = 5000 \times 10 = 50000\, \text{J} = 5\times10^{4}\, \text{J} \) per piano.

Now, if the lunch energy is, say, \( E = 2.5\times10^{6}\, \text{J} \) (common value for a lunch's energy), then number of pianos \( n = \frac{E}{W} \).

Wait, but the problem says "Complete the energy bar graph", but maybe the energy from lunch is given in the original (maybe part a, not shown). Assuming the lunch energy is \( 2.5\times10^{6}\, \text{J} \) (typical), then:

Step1: Calculate work per piano

Work to lift one piano: \( W = F \times d \)
\( W = 5000\, \text{N} \times 10\, \text{m} = 50000\, \text{J} \)

Step2: Assume lunch energy (if not given, but common lunch energy is \( 2.5\times10^{6}\, \text{J} \))

Let \( E_{\text{lunch}} = 2.5\times10^{6}\, \text{J} \) (from typical lunch, or from part a of the problem)

Step3: Calculate number of pianos

\( n = \frac{E_{\text{lunch}}}{W} = \frac{2.5\times10^{6}\, \text{J}}{5\times10^{4}\, \text{J}} = 50 \)

Wait, but maybe the lunch energy is different. Wait, the problem might have a missing value. Wait, the user's problem as shown has a typo or missing data? Wait, no, maybe in the original problem (the image), the energy from lunch is given. Since the user provided the problem, maybe there's a missing part. But assuming the lunch energy is \( 2.5\times10^{6}\, \text{J} \) (common), then:

Answer:

If lunch energy is \( 2.5\times10^{6}\, \text{J} \), number of pianos is \( \boldsymbol{50} \).

(Note: If the lunch energy is different, recalculate with \( n = \frac{E_{\text{lunch}}}{50000} \). For example, if lunch energy is \( 1\times10^{6}\, \text{J} \), then \( n = 20 \), etc. The key is work per piano is \( 5\times10^{4}\, \text{J} \), then divide total lunch energy by that.)

Wait, maybe the lunch energy is from a previous part (part a) where, say, the student's lunch energy is \( 2.5 \times 10^6 \) J (like 600 kcal, converted: 600 kcal = 600×4184 J ≈ 2.51×10^6 J). So using that:

Number of pianos \( n = \frac{2.5\times10^6}{5\times10^4} = 50 \).

So the answer is 50 (assuming lunch energy is \( 2.5\times10^6 \) J).