Problem 17 - An Introduction to Mechanics

Question

A \(55-\mathrm{kg}\) athlete leaps into the air from a crouching position. Her center of mass rises \(60 \mathrm{~cm}\) as her feet leave the ground and then it continues another \(80 \mathrm{~cm}\) to the top of the leap. What is the average power she develops, assuming the force on the ground is constant?

Step-by-Step Solution

Answer

Assuming a leap duration of 0.5 seconds, the average power is 1509.2 W.

Step 1: Understanding the Problem

The problem requires us to find the average power developed by the athlete during her leap. Power is the rate at which work is done. The total work done by the athlete can be considered as the work done against gravity to raise her center of mass. The energy required can be calculated using gravitational potential energy.

Step 2: Calculate the Total Height Raised

The athlete’s center of mass rises a total of 60 cm initially and an additional 80 cm, for a total height rise of 140 cm. Convert the total height rise to meters: \[60 \, \text{cm} + 80 \, \text{cm} = 140 \, \text{cm} = 1.4 \, \text{m}\]

Step 3: Calculate the Work Done

Calculate the work done against gravity by using the formula for gravitational potential energy: \[W = mgh\]where \(m = 55 \mathrm{~kg}\), \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity), and \(h = 1.4 \, \text{m}\). Substituting the values, we get:\[W = 55 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.4 \, \text{m} = 754.6 \, \text{J}\]

Step 4: Assume Time Duration for Leap

To calculate average power, we also need the time duration of the leap. Since not provided, we need to assume or infer it. For typical jump activity, assume a time duration. Sometimes, \(t = 0.5\) seconds is a reasonable estimate for a complete leap.

Step 5: Calculate the Average Power Developed

Average power is calculated by dividing the total work done by the time interval:\[P_{avg} = \frac{W}{t}\]With \(W = 754.6 \, \text{J}\) and assuming \(t = 0.5 \, \text{s}\), the average power is:\[P_{avg} = \frac{754.6 \, \text{J}}{0.5 \, \text{s}} = 1509.2 \, \text{W}\]

Step 6: Verification and Conclusion

The calculated average power is reasonable for a short burst of muscular activity like a jump. In conclusion, assuming a leap time of about 0.5 seconds, the average power developed is approximately 1509.2 W.

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is essentially the energy an object holds due to its position in a gravitational field. When an athlete leaps into the air, her center of mass moves upwards, gaining height. This height gain correlates directly to an increase in gravitational potential energy. To calculate this energy increase, we use the formula: \[ W = mgh \] where:

• \( W \) is the work done and hence the energy in joules (J).
• \( m \) is the mass of the athlete (55 kg in this case).
• \( g \) is the gravitational constant, approximately \( 9.8 \, \text{m/s}^2 \).
• \( h \) is the total height change, converted into meters (1.4 m here).

In the given problem, we calculated that the gravitational potential energy gained by the athlete is 754.6 J. This energy is a result of the athlete exerting force against gravity to raise her center of mass.

Power Calculation

Power in physics is the rate at which work is done or energy is transferred. In simpler terms, it tells us how quickly energy is used. In the case of the jumping athlete, once we have determined the amount of work done (in this case lifting her body against gravity), we need to determine how fast this work was done. This is where power comes in. The formula for calculating power is: \[ P = \frac{W}{t} \] where:

• \( P \) stands for power, measured in watts (W).
• \( W \) is the work done, here calculated as 754.6 J.
• \( t \) is the time taken to perform the work.Typically for jumps, we assume a short time duration like 0.5 s.

In our problem, using the aforementioned duration, the athlete's power was calculated to be approximately 1509.2 W. This high power level illustrates the rapid energy expenditure typical in human jumps.

Work Done Against Gravity

Work done against gravity involves moving an object in the opposite direction of gravitational pull. Whenever an athlete jumps or lifts an object, they fight against gravity to increase altitude or height. This opposition requires energy, which is the 'work done' in this context.The formula \( W = mgh \) helps us quantify this work, as it measures the amount of energy required to lift a mass \( m \) over a height \( h \), with gravity \( g \) exerting a force back down. For an athlete weighing 55 kg, jumping a total height of 1.4 m, this results in work done of 754.6 J.This concept highlights that any vertical movement against gravity involves converting kinetic energy into gravitational potential energy, emphasizing both energy conservation and transfer in physical activities.