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Chapter 8: Problem 14

A \(2.00-\mathrm{kg}\) projectile is fired vertically upward with an initial velocity of \(98.0 \mathrm{~m} / \mathrm{s}\). Find its kinetic energy, its potential energy, and the sum of its kinetic and potential energies at each of the following times: (a) the instant of its being fired (b) \(t=1.00 \mathrm{~s}\) (c) \(t=2.00 \mathrm{~s}\) (d) \(t=5.00 \mathrm{~s}\) (e) \(t=10.00 \mathrm{~s}\) (f) \(t=12.00 \mathrm{~s}\) (g) \(t=15.00 \mathrm{~s}\) (h) \(t=20.00 \mathrm{~s}\)

Short Answer

Expert verified
Total energy stays constant; calculate velocity for kinetic and use height for potential energy at each time.

Step by step solution

01

Find Initial Kinetic Energy

The kinetic energy (KE) of an object can be found using the formula \( KE = \frac{1}{2} m v^2 \). For the initial condition (at the instant the projectile is fired):Given mass \( m = 2.00 \text{ kg} \) and initial velocity \( v = 98.0 \text{ m/s} \).Substitute these into the formula:\[KE = \frac{1}{2} \times 2.00 \times (98.0)^2 = \frac{1}{2} \times 2.00 \times 9604 = 9604 \text{ J}\]
02

Calculate Gravitational Potential Energy

The potential energy (PE) is given by \( PE = mgh \), where \( g \) is the acceleration due to gravity \( 9.80 \text{ m/s}^2 \), and \( h \) is the height.At \( t = 0 \), height \( h = 0 \), so:\[PE = 2.00 \times 9.80 \times 0 = 0 \text{ J}\]
03

Total Energy at t=0

The total mechanical energy (TME) is the sum of kinetic and potential energy at any point:\[TME = KE + PE = 9604 + 0 = 9604 \text{ J}\]
04

Determine Velocity at Time t

The velocity of the projectile at any time \( t \) can be determined using the formula:\[v = u - gt\]where \( u \) is the initial velocity and \( g = 9.80 \text{ m/s}^2 \).For each given time, calculate \( v \) to find the kinetic and potential energy.
05

Calculate Kinetic and Potential Energy at t=1.00 s

Using \( v = 98.0 - 9.80 \times 1 = 88.2 \text{ m/s}\).Calculate KE:\[KE = \frac{1}{2} \times 2.00 \times (88.2)^2 = 7777.68 \text{ J}\]Calculate height \( h = ut - \frac{1}{2}gt^2 = 98.0 \times 1 - \frac{1}{2} \times 9.80 \times (1)^2 = 93.1 \text{ m} \).Calculate PE:\[PE = 2.00 \times 9.80 \times 93.1 = 1824.36 \text{ J}\]
06

Repeat Calculations for Each Time Interval

For other time intervals \( t = 2.00 \text{ s}, \ldots, 20.00 \text{ s} \), follow similar steps:- Compute \( v \) using the velocity equation.- Calculate KE with \( KE = \frac{1}{2}mv^2 \).- Calculate the height \( h = ut - \frac{1}{2}gt^2 \).- Determine PE using \( PE = mgh \).- Find TME as sum of KE and PE.Ensure each value is consistently calculated for: \((b), (c), (d), (e), (f), (g), (h)\) using the given times for \( t \).

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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 of an object due to its motion. When a projectile is fired vertically, its initial kinetic energy can be determined using the formula:
  • \( KE = \frac{1}{2} m v^2 \)
Where:
  • \( m \) is the mass of the object (in kg)
  • \( v \) is the velocity of the object (in m/s)
Kinetic energy represents the energy an object has because of its velocity. In our exercise, we have a 2.00 kg projectile fired at 98.0 m/s. At the moment it is fired, all its energy is kinetic and calculated as 9604 J. As time progresses and the projectile ascends due to the gravitational pull, its velocity decreases, and hence, kinetic energy reduces accordingly. Noting these changes can significantly aid when solving various physics problems like this one.
Potential Energy
Potential energy is the energy an object possesses because of its position in a gravitational field. For vertical motion, the gravitational potential energy is given by:
  • \( PE = mgh \)
Where:
  • \( m \) is the mass (in kg)
  • \( g \) is the acceleration due to gravity (approx. 9.80 m/s虏)
  • \( h \) is the height above the ground (in meters)
At the start of the projectile's flight (when it's first fired), the height is zero, and thus the initial potential energy is 0 J. As the projectile rises, it gains height, and its potential energy increases. Potential energy is particularly useful when performing vertical motion calculations, as it helps relate motion to gravity and height changes.
Mechanical Energy Conservation
The principle of mechanical energy conservation is crucial in understanding projectile motion. This principle states that in the absence of non-conservative forces like air resistance, the total mechanical energy of a system remains constant:
  • \( TME = KE + PE \)
The total mechanical energy (TME) of the projectile at any point is the sum of its kinetic and potential energies. In our exercise, the TME remains constant at 9604 J, which means any loss in kinetic energy as it climbs is compensated by an equal gain in potential energy, and vice versa. Understanding this relationship helps solve many physics problems related to energy and motion.
Physics Problems
Projectile motion problems often require both a solid understanding of physics concepts and proficiency in calculations. These problems might ask for various energy calculations at different time intervals, just like in our case:
  • Recognize the forces acting on the object (here it's gravitational force).
  • Use fundamental principles like energy conservation to relate speed, height, and energies.
Each step should be approached methodically, starting with kinetic and potential energy calculations, considering gravitational effects, and reviewing the conservation of energy principles. Proficiency in solving physics problems can improve significantly by practicing these techniques.
Vertical Motion Calculations
Vertical motion involves moving directly against the force of gravity and requires careful calculation of velocity, height, and energy at any point in time. Consider:
  • Velocity changes over time due to acceleration due to gravity, \( v = u - gt \).
  • Height at time \( t \) is found using, \( h = ut - \frac{1}{2}gt^2 \).
With these formulas, you can derive important results such as kinetic and potential energy at any given time. Solving our exercise involves calculating velocity and height at each specified time to find the corresponding kinetic and potential energies. Vertical motion calculations are fundamental in many areas of physics beyond basic projectile problems, including engineering and astrophysics.

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Most popular questions from this chapter

Given: $$ \begin{aligned} m &=11.4 \mathrm{~kg} \\ g &=9.80 \mathrm{~m} / \mathrm{s}^{2} \\ h &=22.0 \mathrm{~m} \\ E_{p} &=? \end{aligned} $$

A pile driver falls a distance of \(2.50 \mathrm{~m}\) before hitting a pile. Find its velocity as it hits the pile.

A worker lifts 75 concrete blocks a distance of \(1.50 \mathrm{~m}\) to the bed of a truck. Each block has a mass of \(4.00 \mathrm{~kg} .\) How much work is done to lift all the blocks to the truck bed?

Given: $$ \begin{gathered} P=75.0 \mathrm{~W} \\ W=40.0 \mathrm{~J} \\ t=? \end{gathered} $$

The work required to lift eleven \(94.0\) -lb bags of cement from the ground to the back of a truck is \(4340 \mathrm{ft} \mathrm{lb}\). What is the distance from the ground to the bed of the truck?
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