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A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising the helicopter. An \(810-\mathrm{kg}\) helicopter rises from rest to a speed of \(7.0 \mathrm{m} / \mathrm{s}\) in a time of \(3.5 \mathrm{s}\) During this time it climbs to a height of \(8.2 \mathrm{m}\). What is the average power generated by the lifting force?

Short Answer

Expert verified
The average power is approximately 24302.2 watts.

Step by step solution

01

Calculate the Kinetic Energy

To find the average power, we need to calculate the work done. First, we calculate the kinetic energy gained by the helicopter. The formula for kinetic energy (KE) is:\[ KE = \frac{1}{2} m v^2 \]where \( m = 810 \, \text{kg} \) is the mass, and \( v = 7.0 \, \text{m/s} \) is the final velocity. Plugging in the values:\[ KE = \frac{1}{2} \times 810 \, \text{kg} \times (7.0 \, \text{m/s})^2 = 19845 \, \text{J} \].
02

Calculate the Gravitational Potential Energy

Next, calculate the gravitational potential energy (GPE) gained by the helicopter. The formula for gravitational potential energy is:\[ GPE = mgh \]where \( m = 810 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 8.2 \, \text{m} \) is the height risen. Plugging in the values:\[ GPE = 810 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 8.2 \, \text{m} = 65211.6 \, \text{J} \].
03

Calculate Total Work Done

The total work done by the lifting force is the sum of the kinetic energy and the gravitational potential energy:\[ \text{Total Work} = KE + GPE = 19845 \, \text{J} + 65211.6 \, \text{J} = 85056.6 \, \text{J} \].
04

Calculate Average Power

The average power generated is the total work done divided by the time. The formula for power \( P \) is:\[ P = \frac{\text{Work}}{\text{time}} \]Given that the time \( t = 3.5 \, \text{s} \), we can calculate:\[ P = \frac{85056.6 \, \text{J}}{3.5 \, \text{s}} = 24302.2 \, \text{W} \].
05

Conclusion

The average power generated by the lifting force is approximately 24302.2 watts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. Whenever an object is in motion, it has kinetic energy. It is an essential concept in physics because it relates directly to the movement of objects. The formula for kinetic energy is given by:
  • \[ KE = \frac{1}{2} m v^2 \]
Here, \( m \) is the mass of the object, and \( v \) is its velocity. In the context of our helicopter exercise, the helicopter gained kinetic energy as it accelerated from rest to a speed of 7.0 m/s. This transition from a stationary position to motion builds up kinetic energy.
Calculating kinetic energy helps us understand the increase in energy as the speed of the helicopter rises. By plugging the given mass and velocity into the equation, we can find out how much energy the helicopter gained solely from its motion. This analysis is crucial for evaluating the energy transitions involved in the helicopter's ascent.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its position relative to the Earth. Any time an object is raised to a height, it gains gravitational potential energy. This energy depends on three factors:
  • The mass of the object (\( m \))
  • The height (\( h \)) above the ground
  • The acceleration due to gravity (\( g \)), approximately \( 9.8 \, \text{m/s}^2 \)

The formula for calculating gravitational potential energy is:
  • \[ GPE = mgh \]
In our helicopter scenario, as it climbed to a height of 8.2 meters, it gained gravitational potential energy. This energy represents the work done against the gravitational force to lift the helicopter to its new altitude. Gravitational potential energy is crucial for understanding how much work is involved in raising the helicopter. By calculating GPE, we ascertain the energy gain due to the helicopter's change in position.
Work Done
Work done is a term that describes the amount of energy transferred by a force acting on an object as it moves through a distance. In physics, work is calculated using the equation:
  • \[ \text{Work} = \text{Force} \times \text{Distance} \]
Work is particularly significant when it comes to lifting objects vertically against gravitational forces, as seen with our helicopter. In this case, the work done on the helicopter is the total kinetic energy and gravitational potential energy gained during its ascent.
In exercises like the helicopter problem, calculating work reveals the overall energy required to move the helicopter against both inertia and gravity. This comprehensive approach allows us to determine the total energy expenditure, which is essential for computing the average power generated by the lifting force. Understanding work done summarizes how different energy forms collectively enable the helicopter's motion and altitude gain.

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Most popular questions from this chapter

A \(1.00 \times 10^{2}-\mathrm{kg}\) crate is being pushed across a horizontal floor by a force \(\overrightarrow{\mathbf{P}}\) that makes an angle of \(30.0^{\circ}\) below the horizontal. The coefficient of kinetic friction is \(0.200 .\) What should be the magnitude of \(\overrightarrow{\mathbf{P}},\) so that the net work done by it and the kinetic frictional force is zero?

A cable lifts a 1200-kg elevator at a constant velocity for a distance of \(35 \mathrm{m} .\) What is the work done by (a) the tension in the cable and (b) the elevator's weight?

Starting from rest, a \(1.9 \times 10^{-4}-\mathrm{kg}\) flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of \(0.38 \mathrm{N}\) on it. This force does \(+2.4 \times 10^{-4} \mathrm{J}\) of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.

A \(55-\mathrm{kg}\) box is being pushed a distance of \(7.0 \mathrm{m}\) across the floor by a force \(\overrightarrow{\mathbf{P}}\) whose magnitude is \(160 \mathrm{N}\). The force \(\overrightarrow{\mathbf{P}}\) is parallel to the displacement of the box. The coefficient of kinetic friction is \(0.25 .\) Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped \(126 \mathrm{m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed at the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with a speed of \(40.5 \mathrm{m} / \mathrm{s}\) (about \(91 \mathrm{mi} / \mathrm{h}\) ). How much work was done on him and his sled (assuming a total mass of \(118 \mathrm{kg}\) ) by nonconservative forces during this heat?

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