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A projectile of mass 0.750 kg is shot straight up with an initial speed of \(18.0 \mathrm{m} / \mathrm{s}\). (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only \(11.8 \mathrm{m},\) determine the magnitude of the average force due to air resistance.

Short Answer

Expert verified
(a) 16.53 m, (b) 2.92 N

Step by step solution

01

Identify the Known Variables

The mass of the projectile is given as \( m = 0.750 \) kg, the initial speed as \( v_i = 18.0 \) m/s, and the maximum height reached with air resistance as \( h = 11.8 \) m. Gravity \( g = 9.8 \) m/s² is a constant.
02

Calculate Maximum Height Without Air Resistance

Use the formula for the motion of a projectile without air resistance: \( v_f^2 = v_i^2 - 2gh' \) where \( v_f = 0 \) and solve for \( h' \).\[ \begin{align*} 0 &= (18.0)^2 - 2 \times 9.8 \times h' \ 2 \times 9.8 \times h' &= 324 \ h' &= \frac{324}{19.6} \approx 16.53 \text{ m} \end{align*} \]
03

Calculate Work Done by Air Resistance

Calculate the work done against air resistance using the energy difference. The difference in potential energy is \( mgh' - mgh = 0.750 \times 9.8 \times (16.53 - 11.8) \approx 34.51 \text{ J} \). This work is done by air resistance.
04

Find Average Force of Air Resistance

The work-energy principle states \( W = F_{average} \times d \), where \( W \) is the work done by the air resistance, and \( d = h \) is the distance over which the force acts.\[ F_{average} = \frac{34.51}{11.8} \approx 2.92 \text{ N} \]

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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. It depends on two key factors: the mass of the object and its speed. The formula to calculate kinetic energy (KE) is given by \(KE = \frac{1}{2}mv^2\), where \(m\)is the mass in kilograms, and \(v\)is the velocity in meters per second. In our example, the projectile has an initial velocity of 18.0 m/sand a mass of 0.750 kg. When it was launched, it had a kinetic energy calculated as follows:
  • \( KE = \frac{1}{2} (0.750 \, \text{kg})(18.0 \, \text{m/s})^2\)
  • This gives a kinetic energy of \(121.5 \, \text{Joules}\)just as it leaves the ground.
As the projectile moves upwards, its kinetic energy transforms into potential energy. This transformation continues until the projectile reaches the peak of its trajectory, where its speed, and therefore its kinetic energy, is zero.
Potential Energy
Potential energy is energy that is stored within an object due to its position or height. For an object lifted against gravity, its potential energy can be calculated using the formula:\(PE = mgh\),where \(m\)is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height above a reference point.In the projectile motion exercise, as the projectile rises, its kinetic energy is converted to potential energy. When the projectile reaches its highest point, its potential energy is at maximum:
  • Without air resistance, using height 16.53 m,
  • Potential Energy would be \(PE = 0.750 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 16.53 \, \text{m} \approx 121.5 \, \text{J}\).
This conversion happens because all the kinetic energy initially imparted is used to overcome gravitational potential energy as height increases.
Air Resistance
Air resistance is a force that opposes the motion of an object through the air. It acts in the opposite direction to the object's movement and can significantly affect the motion of objects, especially at high speeds or with large surface areas. In our projectile problem, air resistance is the reason the projectile rises to only 11.8 m instead of reaching 16.53 m. Some key points about air resistance:
  • It depends on factors such as the object's velocity, cross-sectional area, air density, and the drag coefficient.
  • Higher speed means greater air resistance.
  • In our case, air resistance causes a difference in how high the projectile goes compared to if it were in a vacuum.
Air resistance converts some of the projectile's kinetic energy to other forms of energy, like thermal energy, preventing the projectile from achieving its full potential height.
Work-Energy Principle
The work-energy principle links the concepts of work and energy. It states that the work done on an object equals the change in its kinetic energy. This principle can also be applied to potential energy changes, especially in projectile motion where energy conservation is a key concept.In our exercise, after calculating the potential energy difference:
  • When no air resistance is present, potential energy at maximum height would be \(121.5 \, \text{J}\).
  • With air resistance, potential energy is \(87.0 \, \text{J}\),using the height 11.8 m.
The difference of 34.5 Jis work done by air resistance:
  • Using the equation \(W = F_{average} \times d\), where \(W\)is work done and \(d\)is distance over which the force is applied.
  • This allows us to find the average force of air resistance using \(F_{average} = \frac{34.5 \text{J}}{11.8 \text{m}} \approx 2.92 \, \text{N}\).
Understanding this principle helps to connect work done against forces and energy transformations during movements like projectile motion.

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Most popular questions from this chapter

As a sailboat sails 52 m due north, a breeze exerts a constant force \(\overrightarrow{\mathbf{F}}_{1}\) on the boat's sails. This force is directed at an angle west of due north. A force \(\overrightarrow{\mathbf{F}}_{2}\) of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just \(47 \mathrm{m} .\) What is the angle between the direction of the force \(\overrightarrow{\mathbf{F}}_{1}\) and due north?

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by seven-time-winner Lance Armstrong \((m=75.0 \mathrm{kg})\) is \(6.50 \mathrm{W}\) per kilogram of his body mass. (a) How much work does he do during a 135 -km race in which his average speed is \(12.0 \mathrm{m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

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?

A car, starting from rest, accelerates in the \(+x\) direction as in the figure. It has a mass of \(1.10 \times 10^{3} \mathrm{kg}\) and maintains an acceleration of \(+4.60 \mathrm{m} / \mathrm{s}^{2}\) for \(5.00 \mathrm{s}\). Assume that a single horizontal force (not shown) accelerates the vehicle. Determine the average power generated by this force.

A basketball player makes a jump shot. The \(0.600-\mathrm{kg}\) ball is released at a height of \(2.00 \mathrm{m}\) above the floor with a speed of \(7.20 \mathrm{m} / \mathrm{s}\). The ball goes through the net \(3.10 \mathrm{m}\) above the floor at a speed of \(4.20 \mathrm{m} / \mathrm{s}\). What is the work done on the ball by air resistance, a nonconservative force?

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