Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 17

A pole-vaulter of mass \(60.0 \mathrm{kg}\) vaults to a height of $6.0 \mathrm{m}$ before dropping to thick padding placed below to cushion her fall. (a) Find the speed with which she lands. (b) If the padding brings her to a stop in a time of \(0.50 \mathrm{s},\) what is the average force on her body due to the padding during that time interval?

Short Answer

Expert verified
Question: Calculate the landing speed of a 60 kg pole-vaulter who reaches a height of 6 meters, and find the average force exerted on her body by the padding during the time interval of 0.5 seconds as she stops. Answer: The landing speed is approximately 9.6 m/s, and the average force exerted on her body is around -1152 N.

Step by step solution

01

Find the potential energy at the highest point of the vault

The pole-vaulter's potential energy at the highest point is given by: \(PE = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. Plugging in the given values to the equation, we get: \(PE = (60.0\:\text{kg})(9.81\:\text{m/s}^2)(6.0\:\text{m})\).
02

Calculate the kinetic energy at the landing point

Since mechanical energy is conserved, the potential energy at the highest point is equal to the kinetic energy at the landing point. Hence, \(KE = PE\). We have already found the value of \(PE\), so we can directly use it: \(KE = 3531.6\:\text{J}\).
03

Obtain the landing speed

The kinetic energy is given by the equation: \(KE=\dfrac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the speed. Solving for \(v\), we get: \(v = \sqrt{\dfrac{2 \times KE}{m}}\). Plugging in the values, we find the landing speed: \(v = \sqrt{\dfrac{2 \times 3531.6\:\text{J}}{60.0\:\text{kg}}} \approx 9.6\:\text{m/s}\).
04

Calculate the deceleration experienced due to the padding

Using Newton's second law, we can relate force, mass, and deceleration: \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the deceleration. Rearranging the equation, we get: \(a = \dfrac{F}{m}\). Since we know the time it takes for her to come to a stop, we can find the deceleration using the equation: \(v_f = v_i + at\), where \(v_f\) is the final velocity (0 m/s), \(v_i\) is the initial velocity (9.6 m/s), and \(t\) is the time interval (0.5 s). Rearranging the equation, we get: \(a = \dfrac{v_f - v_i}{t}\). Plugging in the values, we find the deceleration: \(a = \dfrac{0 - 9.6\:\text{m/s}}{0.5\:\text{s}}=-19.2\:\text{m/s}^2\).
05

Determine the average force on her body during the stopping time

Plugging the deceleration value into the equation \(F = ma\), we can calculate the average force exerted on the pole-vaulter: \(F = (60.0\:\text{kg})(-19.2\:\text{m/s}^2)\approx -1152\:\text{N}\). The force is negative because it acts in the opposite direction of the motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 0.15-kg baseball traveling in a horizontal direction with a speed of $20 \mathrm{m} / \mathrm{s}$ hits a bat and is popped straight up with a speed of \(15 \mathrm{m} / \mathrm{s} .\) (a) What is the change in momentum (magnitude and direction) of the baseball? (b) If the bat was in contact with the ball for 50 ms, what was the average force of the bat on the ball?
Two pendulum bobs have equal masses and lengths \((5.1 \mathrm{m}) .\) Bob \(\mathrm{A}\) is initially held horizontally while bob \(\mathrm{B}\) hangs vertically at rest. Bob A is released and collides elastically with bob B. How fast is bob B moving immediately after the collision?
A 0.15-kg baseball is pitched with a speed of \(35 \mathrm{m} / \mathrm{s}\) \((78 \mathrm{mph}) .\) When the ball hits the catcher's glove, the glove moves back by \(5.0 \mathrm{cm}(2 \text { in. })\) as it stops the ball. (a) What was the change in momentum of the baseball? (b) What impulse was applied to the baseball? (c) Assuming a constant acceleration of the ball, what was the average force applied by the catcher's glove?
A \(0.020-\mathrm{kg}\) bullet traveling at \(200.0 \mathrm{m} / \mathrm{s}\) east hits a motionless \(2.0-\mathrm{kg}\) block and bounces off it, retracing its original path with a velocity of \(100.0 \mathrm{m} / \mathrm{s}\) west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.
A radium nucleus (mass 226 u) at rest decays into a radon nucleus (symbol Rn, mass 222 u) and an alpha particle (symbol \(\alpha,\) mass 4 u). (a) Find the ratio of the speeds \(v_{\alpha} / v_{\mathrm{Rn}}\) after the decay. (b) Find the ratio of the magnitudes of the momenta \(p_{\alpha} / p_{\mathrm{Rn}^{*}}\) (c) Find the ratio of the kinetic energies \(K_{\alpha} / K_{\mathrm{Rn}} .\) (Note: "u" is a unit of mass; it is not necessary to convert it to kg.)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Nuclear Physics

Read Explanation

Force

Read Explanation

Mechanics and Materials

Read Explanation

Space Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free