Question

Heedless of danger, a child leaps onto a pile of old mattresses to use them as a trampoline. His motion between two particular points is described by the energy conservation equation

$\frac{\mathbf{1}}{\mathbf{2}}\left(46.0kg\right){\left(2.40m/s\right)}^{\mathbf{2}}\mathbf{+}\left(46.0kg\right)\left(9.80m/s^2\right)\left(2.80m+x\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left(1.94\times {10}^{4}N/m\right){\mathit{x}}^{\mathbf{2}}$

(a) Solve the equation for x. (b) Compose the statement of a problem, including data, for which this equation gives the solution. (c) Add the two values of x obtained in part (a) and divide by 2. (d) What is the significance of the resulting value in part (c)?

Short answer

(a)For the given equation, the values of are found to be 0.403m or -0.357m.

(b) The statement for the problem is given below.

(c) The required value is 0.0232m

(d) The result is the distance by which the mattresses compresses if the child just stands on them. It is the mean point for oscillations produced the matresses.

Step-by-step solution

Step 1:Introduction

In the absence of the non-conservative forces, the total energy of any system is conserved. Thus, the change in total energy is zero.

$\mathbf{∆}{\mathit{E}}_{\mathbf{sys}}\mathbf{=}\mathbf{0}$

Law of conservation of energy is written as

$\mathbf{∆}\mathit{K}\mathbf{+}\mathbf{∆}\mathit{U}\mathbf{=}\mathbf{0}$

The potential energy of the system is

$\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{R}$

The kinetic energy of the system is

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

where

U=Potential energy

K= Kinetic energy

v=Speed of body

g= Gravitational acceleration

R= Displacement in meter

m=Mass of body

Given Data

Given equation is$\frac{1}{2}\left(46.0\text{kg}\right){\left(2.40\text{m/s}\right)}^2+\left(46.0\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\left(2.80\text{m}+x\right)=\frac{1}{2}\left(1.94\times {10}^4\text{N/m}\right)x^2$

calculation for part (a)

Simplifying the equation

$\left(9700\text{N/m}\right)x^2-\left(450.8\text{N}\right)x-1395\text{N.m}=0$

Then

$\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(450.8\text{N}\right)\pm \sqrt{{\left(450.8\text{N}\right)}^2-4\left(9700\text{N/m}\right)\left(-1395\text{N.m}\right)}}{2\left(9700\text{N/m}\right)}\\ =\frac{450.8\text{N}\pm 7357\text{N}}{19400\text{N/m}}=0.403\text{m}\text{or}-0.357\text{m}\end{array}$

calculation for part (b)

From a perch at a height of above the top of a pile of mattresses, a child jumps upward at and lands on the pile of mattresses. The mattresses act as a linear spring with force constant. Find the compression in mattresses when child lands on them.

calculation for part (c)

$\frac{0.403m+\left(-0.357m\right)}{2}=0.0232m$

calculation for part (d)

This result is compression produced in the mattresses if the child just stands on them. It represents the mean position for the oscillations produced in the mattress.