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Chapter 5: Problem 19

Jumping into snow. Albert jumps from the roof of his house into a deep pile of snow. He starts \(5 \mathrm{~m}\) above the snow, and stops \(1 \mathrm{~m}\) into the snow. What force was exerted on Albert? Albert's mass is \(m=70 \mathrm{~kg}\). You can ignore air resistance, and you can assume that the force on Albert from the snow is constant.

Short Answer

Expert verified
The force exerted on Albert is 3430 N.

Step by step solution

01

- Determine the potential energy

Calculate Albert's gravitational potential energy just before he hits the snow using the formula: \[ PE = m \times g \times h \]where:- \(m\) is the mass (70 \mathrm{~kg})- \(g\) is the acceleration due to gravity (9.8 \mathrm{~m/s^2})- \(h\) is the height of the fall (5 \mathrm{~m})Substituting the values: \[ PE = 70 \times 9.8 \times 5 = 3430 \text{ J} \]
02

- Relate work done to potential energy

The work done by the snow to stop Albert is equal to the potential energy he had before hitting the snow. Therefore, \( W = 3430 \text{ J} \).
03

- Calculate the stopping distance

The stopping distance is the depth into the snow that Albert goes, i.e., \( d = 1 \text{ m} \).
04

- Apply the work-energy principle

The work done is also given by: \[ W = F \times d \]where \( F \) is the force exerted by the snow. Rearrange this to solve for F: \[ F = \frac{W}{d} \]
05

- Solve for the force

Substitute the known values into the equation: \[ F = \frac{3430}{1} = 3430 \text{ N} \]

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Key Concepts

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

Potential Energy
Potential energy is the energy stored in an object because of its position relative to some reference point. In Albert's case, his potential energy just before he jumps is determined by his height above the snow. The formula to calculate potential energy is given by: \[ PE = m \times g \times h \]In this formula: \
    \
  • m represents mass
    • \
  • g stands for the acceleration due to gravity (9.8 m/s²)
    • \
  • h is the height
    • \
For Albert, who weighs 70 kg and is 5 meters above the snow, substituting the values into the formula gives us a potential energy of 3430 Joules. This energy is the amount Albert has just before he jumps.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The principle links potential energy to work. When Albert falls into the snow, his potential energy (3430 Joules) transforms into the work done by the snow to stop him.
Mathematically, the work done \[\ W \ = \ F \times d \]where F is the force exerted by the snow and d is the stopping distance. In Albert's case, knowing that the stopping distance is 1 meter, we can solve for the force.
The principle helps us appreciate how energy transformations happen in different scenarios, showing directly how potential energy converts into work done on Albert by the snow.
Gravitational Force
Gravitational force is the attractive force between two masses. On Earth, this force depends on both the mass of the object and the acceleration due to gravity. Its formula is: \[ F = m \times g \]Here \m\ is the mass, and \g\ is the gravitational acceleration (9.8 m/s² on Earth)
In Albert's case, we can calculate the gravitational force he experiences using his mass: \[ F = 70 \times 9.8 \ = 686 \text{ N} \]Understanding gravitational force helps students see how Albert's mass and gravity work together to give him this force. This force also influences the potential energy he has before he jumps.
Constant Force
A constant force is one that remains the same in magnitude and direction over the distance it acts. In the case of Albert jumping into the snow, we assume the force exerted by the snow remains consistent while stopping him.
To find this force, we rearrange the work formula: \[ W = F \times d \]Solving for F, we get: \[ F = \frac{W}{d} \]Substituting Albert's work (3430 Joules) and the stopping distance (1 meter), the force comes out to be 3430 Newtons. This force shows how the snow interacts with Albert to bring him to a stop evenly, illustrating the concept of a constant force in physics.

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Most popular questions from this chapter

Seat belts. Explain why seat belts reduce the risk of injury if you are involved in a car accident. How would you improve seatbelt design?

Space shuttle with air resistance. During lift-off of the space shuttle the engines provide a force of 35 million Newtons. The mass of the shuttle is approximately 2 million \(\mathrm{kg}\). (a) Draw a free-body diagram of the space shuttle immediately after lift-off. (b) Find an expression for the acceleration of the space shuttle immediately after lift-off. Let us assume that the force from the engines is constant, and that the mass of the space shuttle does not change significantly over the first \(20 \mathrm{~s}\). (c) Find the velocity and position of the space shuttle after \(20 \mathrm{~s}\) if you ignore air resistance. Let us assume that we can describe the air resistance force on the space shuttle with a square law, \(F=-D v|v|\), where \(D \simeq 388 \mathrm{~kg} / \mathrm{m} .\) (d) Develop a program to find the velocity and position of the space shuttle using numerical methods. (e) Find the velocity and position of the space shuttle after \(20 \mathrm{~s}\) if you include air resistance. (f) Plot the velocity and position and compare with the results without air resistance. Comment on the results. Notice that \(D\) depends on the density of the surrounding air, and the density falls when as the shuttle ascends, hence \(D\) actually depends on the height of the shuttle.

In the army. You are told by a friend in the army that the force you feel when you fire a gun is the same as the force felt by a sandbag hit by the bullet because the two forces are action-reactions pairs. Is this true?

Reaction time. Your reaction time can be measured with the help of a fried using a ruler. Your friend holds the ruler vertically between your thumb and index finger. When he releases the ruler, you grab it as soon as you can. If the ruler is placed with the \(0 \mathrm{~cm}\) mark initially between your fingers, how can you use how far the ruler has fallen to find your reaction time? You can assume that you use negligible time to actually grab the ruler as soon as you start moving your finger. (a) Draw a free-body diagram for the ruler when it is falling. (b) Find the position of the ruler as a function of time. (c) Find your reaction time, if the ruler fell a vertical distance \(h\) before you grabbed it. (d) If you are driving in your car at \(80 \mathrm{~km} / \mathrm{h}\), how far do you travel during your reaction time?

Modelling Bungee Jumping Numerically. In this exercise we will study a person bungee jumping. The bungee cord acts as an ideal spring with a spring constant \(k\) when it is stretched, but it has no strength when pushed together. The cord's equilibrium length is \(d\). There is also a form of dampening in the cord, which we will model as a force which is dependent on the speed of the cord's deformation. When the cord is stretched a length \(x\), and is being stretched with the instantaneous speed \(v\), the force from the spring is given as $$ F(x, v)=\left\\{\begin{array}{cl} -k(x-d)-c_{v} v & \text { when } x>d \\ 0 & \text { when } x \leq d \end{array}\right. $$ where \(c_{v}\) is a constant that describes the dampening in the cord, and \(k\) is the spring constant. We set \(x=0\) to be where the bungee cord is attached and let the positive direction of the \(x\)-axis point downwards. A person with a mass \(m\) places the cord around the waist and jumps from the point where it is attached. The initial velocity is \(v_{0}=0\). You can neglect air resistance and assume that the bungee cord is massless. The motion is solely vertical. The acceleration of gravity is \(g\). (a) Draw a free-body diagram of the person when the bungee cord is taut. Name all the forces. (b) At what height is the person hanging when the motion has stopped? (c) Write a numerical algorithm that finds the persons position and velocity at the time \(t+\Delta t\) given the persons position and velocity at a time \(t\). And implement this algorithm in a program that finds the motion of a person bungee jumping. (d) Use your program to plot the height as a function of time, \(x(t)\), for a person of mass \(m=70 \mathrm{~kg}\) jumping with a bungee cord of equilibrium length \(d=20 \mathrm{~m}\) and spring constant \(k=150 \mathrm{~N} / \mathrm{m}\), for \(T=60 \mathrm{~s}\) with a timestep of \(d t=0.01 \mathrm{~s} .\) The acceleration of gravity is \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\). What is a reasonable choice for \(c_{v}\) ? Explain your choice. (e) Is the system conservative during the whole motion, parts of the motion, or not at all? Explain. (f) How would our model be different if we included air resistance?
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