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Chapter 6: Problem 34

"Rocket Man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and accelerates straight upward. At a height of \(16 \mathrm{m},\) his speed is \(5.0 \mathrm{m} / \mathrm{s} .\) His mass, including the propulsion unit, has the approximately constant value of 136 kg. Find the work done by the force generated by the propulsion unit.

Short Answer

Expert verified
The work done by the propulsion unit is 22914.4 J.

Step by step solution

01

Identify Known Values and Formula for Work-Energy

First, note the known values: - Initial speed \( v_i = 0 \; \text{m/s} \)- Final speed \( v_f = 5.0 \; \text{m/s} \)- Height \( h = 16 \; \text{m} \)- Mass \( m = 136 \; \text{kg} \) The work-energy principle states that the work done is equal to the change in mechanical energy of the system. The formula is:\[ W = \Delta KE + \Delta PE \]Where \( \Delta KE \) is the change in kinetic energy and \( \Delta PE \) is the change in potential energy.
02

Calculate Change in Kinetic Energy

The initial kinetic energy \( KE_i \) is given by the formula:\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 136 \; \text{kg} \times (0 \; \text{m/s})^2 = 0 \; \text{J} \]The final kinetic energy \( KE_f \) is:\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 136 \; \text{kg} \times (5.0 \; \text{m/s})^2 = 1700 \; \text{J} \]Therefore, the change in kinetic energy \( \Delta KE \) is:\[ \Delta KE = KE_f - KE_i = 1700 \; \text{J} - 0 \; \text{J} = 1700 \; \text{J} \]
03

Calculate Change in Potential Energy

The initial potential energy \( PE_i \) is \[ PE_i = mgh_i = 136 \; \text{kg} \times 9.8 \; \text{m/s}^2 \times 0 \; \text{m} = 0 \; \text{J} \]The final potential energy \( PE_f \) is \[ PE_f = mgh_f = 136 \; \text{kg} \times 9.8 \; \text{m/s}^2 \times 16 \; \text{m} = 21214.4 \; \text{J} \]Thus, the change in potential energy \( \Delta PE \) is:\[ \Delta PE = PE_f - PE_i = 21214.4 \; \text{J} - 0 \; \text{J} = 21214.4 \; \text{J} \]
04

Calculate Total Work Done

Using the work-energy principle, the total work done \( W \) by the propulsion force is given by the sum of changes in kinetic and potential energies:\[ W = \Delta KE + \Delta PE = 1700 \; \text{J} + 21214.4 \; \text{J} = 22914.4 \; \text{J} \] Thus, the work done by the propulsion unit is \( 22914.4 \; \text{J} \).

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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 the energy an object has due to its motion. Whenever an object is moving, it possesses kinetic energy. This energy depends primarily on two 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 and \( v \) is the velocity.
  • In the context of our rocket exercise, we calculated the kinetic energy twice: once at the start when the speed was zero, and once at a height of 16 meters when the speed was 5 m/s.
  • Initially, the kinetic energy (\( KE_i \)) was zero because the rocket was at rest.
  • Later, with the rocket speeding at \( 5 \, \text{m/s} \), the kinetic energy increased to \( 1700 \, \text{Joules} \).
This change in kinetic energy provides insight into how much energy was transferred to motion. In general, whenever an object speeds up, its kinetic energy increases.
Potential Energy
Potential energy refers to the energy stored within a system due to its position or configuration. In our exercise, this primarily relates to gravitational potential energy.
  • The formula to determine gravitational potential energy (\( PE \)) is: \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2 \) here on Earth), and \( h \) is the height above the ground.
  • Initially, the potential energy (\( PE_i \)) was zero since the rocket started at ground level.
  • At 16 meters high, the potential energy (\( PE_f \)) soared to \( 21214.4 \, \text{Joules} \).
The increase in potential energy explains how much energy was overcome to elevate the rocket against Earth's gravity. Whenever an object is elevated, its potential energy grows.
Rocket Propulsion
Rocket propulsion involves the mechanisms by which rockets and spacecraft move. Propulsion systems generate force that pushes the vehicle forward or upward, overcoming gravitational pull and allowing the vehicle to accelerate.
  • The force generated by the rocket's propulsion unit does the work needed to increase both the kinetic and potential energies of the rocket "Man."
  • This force effectively propels the rocket upwards and against the force of gravity.
  • In our exercise, the propulsion system did a total work of \( 22914.4 \, \text{Joules} \), which is the sum of the changes in kinetic and potential energy.
  • Rocket propulsion systems typically consist of engines that burn fuel to produce thrust, expelling exhaust gases in the opposite direction of intended motion, based on Newton's Third Law of Motion.
Understanding rocket propulsion is key in scenarios where forces must be calculated to ensure targets are reached or maintained, especially in aerospace and astronautics.

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Most popular questions from this chapter

The hammer throw is a track-and-field event in which a \(7.3 \mathrm{kg}\) ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of \(29 \mathrm{m} / \mathrm{s} .\) For comparison, a .22 caliber bullet has a mass of \(2.6 \mathrm{g}\) and, starting from rest, exits the barrel of a gun at a speed of \(410 \mathrm{m} / \mathrm{s}\). Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

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A 35-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 210 J from 440 J. How high does she rise during this interval? Neglect air resistance.

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of \(4.5 \times 10^{4} \mathrm{kg},\) and the force causes its speed to change from 7100 to \(5500 \mathrm{m} / \mathrm{s}\) (a) What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} \mathrm{m},\) determine the magnitude of the force.

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