Alright, let's systematically calculate the force applied by each counterweight and the work done by them using the given formulas.
### Step-by-Step Solution:
1. Given values:
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Distance traveled by each counterweight, [tex]\( d = 5 \, \text{m} \)[/tex]
- Masses of the counterweights:
- [tex]\( 20 \, \text{kg} \)[/tex]
- [tex]\( 100 \, \text{kg} \)[/tex]
- [tex]\( 200 \, \text{kg} \)[/tex]
2. Force applied by each counterweight:
- Formula: [tex]\( F = m \cdot g \)[/tex]
- For the [tex]\( 20 \, \text{kg} \)[/tex] counterweight:
[tex]\[
F_{20} = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N}
\][/tex]
- For the [tex]\( 100 \, \text{kg} \)[/tex] counterweight:
[tex]\[
F_{100} = 100 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 980 \, \text{N}
\][/tex]
- For the [tex]\( 200 \, \text{kg} \)[/tex] counterweight:
[tex]\[
F_{200} = 200 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1960 \, \text{N}
\][/tex]
3. Work done by each counterweight:
- Formula: [tex]\( W = F \cdot d \)[/tex]
- For the [tex]\( 20 \, \text{kg} \)[/tex] counterweight:
[tex]\[
W_{20} = 196 \, \text{N} \times 5 \, \text{m} = 980 \, \text{J}
\][/tex]
- For the [tex]\( 100 \, \text{kg} \)[/tex] counterweight:
[tex]\[
W_{100} = 980 \, \text{N} \times 5 \, \text{m} = 4900 \, \text{J}
\][/tex]
- For the [tex]\( 200 \, \text{kg} \)[/tex] counterweight:
[tex]\[
W_{200} = 1960 \, \text{N} \times 5 \, \text{m} = 9800 \, \text{J}
\][/tex]
### Completed Table 1:
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{Mass of} & \text{Force Applied by} & \text{Work Done by Falling} \\
\text{Counterweight (kg)} & \text{Falling Counterweight} F = m \cdot a & \text{Counterweight} W = F \cdot d \\
\hline
20 & 196 \, \text{N} & 980 \, \text{J} \\
\hline
100 & 980 \, \text{N} & 4900 \, \text{J} \\
\hline
200 & 1960 \, \text{N} & 9800 \, \text{J} \\
\hline
\end{array}
\][/tex]
This table includes all the calculated forces and work done values for each of the counterweights.
