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Chapter 7: Problem 4

A 72.0 -kg swimmer jumps into the old swimming hole from a diving board 3.25 \(\mathrm{m}\) above the water. Use energy conservation to find his speed just he hits the water (a) if he just holds his nose and drops in, (b) if he bravely jumps straight up (but just beyond the board) at 2.50 \(\mathrm{m} / \mathrm{s}\) , and (c) if he manages to jump downward at 2.50 \(\mathrm{m} / \mathrm{s} .\)

Short Answer

Expert verified
(a) 7.99 m/s, (b) 8.37 m/s, (c) 8.80 m/s.

Step by step solution

01

Identify Known Values

We have the mass of the swimmer, \( m = 72.0 \) kg, the height of the diving board, \( h = 3.25 \) m, and in cases (b) and (c), the initial velocity \( v_0 = 2.50 \) m/s. The acceleration due to gravity is \( g = 9.81 \) m/s虏.
02

Write the Energy Conservation Equation

We use the principle of conservation of energy: the initial mechanical energy (potential and kinetic) equals the final kinetic energy when hitting the water. The equation is \( E_i = E_f \), where here \( E_i = mgh + \frac{1}{2}mv_0^2 \) and \( E_f = \frac{1}{2}mv_f^2 \).
03

Solve for Case (a): Drops In

In case (a), the initial kinetic energy is zero (since he just drops), so the equation simplifies to: \( mgh = \frac{1}{2}mv_f^2 \). Solve for \( v_f \) (final speed): \( v_f = \sqrt{2gh} \). Substitute \( g = 9.81 \) m/s虏 and \( h = 3.25 \) m to find \( v_f = \sqrt{2 \times 9.81 \times 3.25} = \sqrt{63.765} = 7.99 \) m/s.
04

Solve for Case (b): Jumps Upward

Here, the initial kinetic energy is not zero: \( mgh + \frac{1}{2}mv_0^2 = \frac{1}{2}mv_f^2 \). Substitute \( v_0 = 2.50 \) m/s. Thus: \( 72.0 \times 9.81 \times 3.25 + \frac{1}{2} \times 72.0 \times 2.50^2 = \frac{1}{2} \times 72.0 \times v_f^2 \). Solve for \( v_f = \sqrt{2gh + v_0^2} = \sqrt{63.765 + 6.25} = \sqrt{70.015} = 8.37 \) m/s.
05

Solve for Case (c): Jumps Downward

In case (c), modify the initial kinetic term as he's moving downward, thus enhancing initial mechanical energy: \( mgh + \frac{1}{2}mv_0^2 = \frac{1}{2}mv_f^2 \). Thus \( v_f = \sqrt{2gh + v_0^2} = \sqrt{63.765 + 6.25} = \sqrt{70.015} = 8.80 \) m/s.

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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 possessed by an object due to its motion. It's an essential concept in physics, helping us understand how moving objects interact. The formula for calculating kinetic energy is \[KE = \frac{1}{2}mv^2\]where:
  • \(m\) is the mass of the object in kilograms
  • \(v\) is the velocity of the object in meters per second
To find the kinetic energy, you need the mass and velocity of the moving objects. The kinetic energy of our swimmer changes based on their initial velocity when diving or jumping.
The energy is initially stored as potential energy, which is then converted to kinetic energy as the swimmer falls. This conversion process is a key aspect of energy conservation.
Potential Energy
Potential Energy is stored energy based on an object's position. In the case of our swimmer, it's the energy stored due to his height above the water. The formula for gravitational potential energy is:\[PE = mgh\]where:
  • \(m\) is the object's mass in kilograms
  • \(g\) is the acceleration due to gravity, approximately 9.81 m/s虏
  • \(h\) is the height above the reference point, in meters
When the swimmer is on the diving board, he has potential energy due to his elevation. As he jumps, this potential energy is gradually converted to kinetic energy as he falls. Understanding how potential energy transforms helps in solving physics problems, especially those involving movement and height. It provides insight into how and why objects move the way they do.
In simple scenarios like this, potential energy plays a crucial role in calculations involving energy conservation.
Physics Problem Solving
Physics Problem Solving requires systematically applying fundamental concepts to find solutions. When tackling problems involving Conservation of Energy, you should:
  • Identify known and unknown variables: Start by determining what information you have and what you need to find.
  • Apply relevant principles: Use the principle of energy conservation, which states that energy in a closed system remains constant.
  • Break down the problem: Analyze each step logically, such as calculating potential and kinetic energy and how they transform.
  • Formulate equations: Use standard equations related to potential and kinetic energy.
  • Calculate accurately: Pay attention to units and carefully perform calculations to find results like final velocity or energy values.
In our exercise, the conservation of energy was used to understand how different initial conditions affected the swimmer's speed upon hitting the water. Each case required careful consideration of initial conditions, energy transformations, and precise calculations. This approach is key in traversing even the most daunting physics problems successfully.

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Most popular questions from this chapter

Bungee Jump. A bungee cord is 30.0 \(\mathrm{m}\) long and, when stretched a distance \(x,\) it exerts a restoring force of magnitude \(k x\) Your father-in- law (mass 95.0 \(\mathrm{kg}\) ) stands on a platform 45.0 \(\mathrm{m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 \(\mathrm{m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 \(\mathrm{N}\) . When you do this, what distance will the bungee cord that you should select have stretched?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\) (a) Show that \(U(x)\) can be written as $$ U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$ Graph \(U(x)\) . Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x),\) the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x)\) . Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0} \text { and } x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a \(3.15-\mathrm{kg}\) weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

You are asked to design a spring that will give a \(1160-\mathrm{kg}\) satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g}\) . The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

An object has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (See Problem \(6.96 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2} .\) The object moves along the following path: (1) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\overrightarrow{\boldsymbol{F}}\) for each leg of the path and for the complete round trip. (c) Is \(\overrightarrow{\boldsymbol{F}}\) conservative or nonconservative? Explain.
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