A girl of weight 440 N climbs up a rope through a height of 7 m in 20 s. The power expended by the girl is:
154 W
This physics problem asks us to calculate the power expended by a girl climbing a rope. Power is the rate at which work is done. In this case, the work done is against the force of gravity as the girl increases her height.
We are given the following information:
The work done (\(W\)) when moving against a constant force over a distance is given by the formula:
\[W = \text{Force} \times \text{Distance}\]
In this scenario, the force is the girl's weight, and the distance is the height she climbs. So, the work done is:
\[W = F \times d\]
Substituting the given values:
\[W = 440 \text{ N} \times 7 \text{ m}\]
\[W = 3080 \text{ Joules}\]
The work done by the girl in climbing the rope is 3080 Joules.
Power (\(P\)) is the rate at which work is done. It is calculated using the formula:
\[P = \frac{\text{Work done}}{\text{Time taken}}\]
Using the work done we just calculated and the time taken:
\[P = \frac{W}{t}\]
Substituting the values:
\[P = \frac{3080 \text{ J}}{20 \text{ s}}\]
\[P = 154 \text{ J/s}\]
Since 1 Joule per second is equal to 1 Watt (W), the power expended by the girl is:
\[P = 154 \text{ W}\]
The power expended by the girl climbing the rope is 154 W.
| Concept | Definition | Formula | SI Unit |
|---|---|---|---|
| Work (W) | Energy transferred by a force acting over a distance. | \(W = F \times d \times \cos(\theta)\) (where \(\theta\) is angle between force and displacement) | Joule (J) |
| Potential Energy (Gravitational) (PE) | Energy stored by an object due to its position in a gravitational field. | \(PE = mgh\) or \(PE = \text{weight} \times \text{height}\) | Joule (J) |
| Power (P) | The rate at which work is done or energy is transferred. | \(P = \frac{W}{t}\) or \(P = F \times v\) (where v is velocity) | Watt (W) |
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