/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A pole-vaulter of mass \(60.0 \m... [FREE SOLUTION] | 91影视

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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.

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