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Bio In a needle biopsy, a narrow strip of tissue is extracted from a patient with a hollow needle. Rather than being pushed by hand, to ensure a clean cut the needle can be fired into the patient's body by a springAssume the needle has mass \(5.60 \mathrm{~g}\), the light spring has force constant \(375 \mathrm{~N} / \mathrm{m}\), and the spring is originally compressed \(8.10 \mathrm{~cm}\) to project the needle horizontally without friction. The tip of the needle then moves through \(2.40 \mathrm{~cm}\) of skin and soft tissue, which exerts a resistive force of \(7.60 \mathrm{~N}\) on it. Next, the needle cuts \(3.50 \mathrm{~cm}\) into an organ, which exerts a backward force of \(9.20 \mathrm{~N}\) on it. Find (a) the maximum speed of the needle and (b) the speed at which a flange on the back end of the needle runs into a stop, set to limit the penetration to \(5.90 \mathrm{~cm}\).

Short Answer

Expert verified
The maximum speed of the needle is approximately \sqrt{(2*375 N/m * 0.081 m^2)/(5.60 g)} m/s. And the speed at which the flange hits the stop is about \sqrt{(2 * (375 N/m * 0.081 m^2 - (7.60 N * 0.024 m + 9.20 N * 0.035 m))/(5.60 g)} m/s.

Step by step solution

01

Calculate the Spring Energy

The spring energy that propels the needle is calculated using the formula for elastic potential energy stored in a compressed spring, which is given as \(E_s=\frac{1}{2} k x^2\), where k is the spring constant and x is the amount of compression. In this case, \(k = 375 N/m\) and \(x = 8.10 cm = 0.081 m\). Substituting these values will give us the initial spring energy.
02

Calculate the Maximum Speed

The maximum speed is acquired when all the spring energy is converted to kinetic energy. We can use the formula for kinetic energy, which is \(KE=\frac{1}{2} m v^2\) where m is the mass of body and v is its velocity. Since the whole spring energy is converted into kinetic energy, we can equate the expressions and isolate the variable v, which is the maximum speed of needle.
03

Calculate Resistive Forces

Next, we need to account for the resistive forces that the needle encounters which cause it to decelerate. The needle experiences a resistive force of $7.60 N$ for \(2.40 cm\) and another $9.20 N$ for \(3.50 cm\). To find the total work done against the resistive forces, it will be helpful to use the formula of work \(W=F \cdot d\) where F is force and d is distance.
04

Calculate Final Speed

The kinetic energy the needle had after leaving the spring is reduced by the total work done against resistive forces and this is converted into kinetic energy for the final stage of the needle's journey. This way, we can find the final speed of the needle. This is accomplished by re-purposing the kinetic energy formula used in Step 2.

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Key Concepts

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

Elastic Potential Energy
When we consider objects like springs, we encounter the concept of elastic potential energy. It's a form of potential energy that is stored when materials stretch or compress. A key example in physics is the spring mechanism involved in a needle biopsy device. The spring is compressed, and as a result, energy is stored in it due to its position.

This stored energy, which can be mathematically expressed as \[\begin{equation}E_s = \frac{1}{2}kx^2\end{equation}\]where \(k\) is the spring constant and \(x\) is the compression distance, can be utilized to do work. In a needle biopsy, this energy propels the needle into tissue. It's crucial for students to understand that this energy is a result of the physical properties of the spring and its deformation (compression in the case of the biopsy needle).
Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. When the spring in our biopsy needle example releases, the stored elastic potential energy is transformed into kinetic energy, propelling the needle forward.

The formula to calculate kinetic energy is \[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\]where \(m\) is the mass and \(v\) is the velocity of the object. Dark.Remember, mass must be in kilograms and velocity in meters per second to use SI units correctly. As the needle moves, its kinetic energy enables it to do work against the resistive forces it encounters.
Work-Energy Principle
The work-energy principle is a foundational concept in physics. It essentially states that work done on an object results in a change in the object's kinetic energy. Work can be calculated using the formula \[\begin{equation}W = F \times d\end{equation}\]where \(F\) is the force applied, and \(d\) is the displacement caused by the force.

In a practical scenario like our needle biopsy, the work done by resistive forces (e.g., skin and organ tissue) is essentially the energy required to overcome the resistance they offer. This work consequently reduces the kinetic energy of the needle, which affects how far it can travel. If the work done by these forces is greater than the initial kinetic energy, the needle will stop before reaching the designed penetration depth.
Resistive Forces
Resistive forces are forces that oppose the motion of an object, often resulting from interactions with other materials or mediums, like air resistance or friction. In medical procedures like a needle biopsy, the needle encounters resistive forces from skin and tissue that work against the motion imparted by the spring.

Calculating the effect of these forces is an application of the work-energy principle. For each segment of the needle's path, the force \[\begin{equation}F\end{equation}\]and the distance \[\begin{equation}d\end{equation}\] it acts over will determine how much work is done against the needle's kinetic energy. It is a good practice to calculate the work done against each resistive force separately, since the forces may vary based on the different tissues the needle passes through.

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Most popular questions from this chapter

Tarzan swings on a \(30.0\)-m-long vine initially inclined at an angle of \(37.0^{\circ}\) with the vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) If he pushes off with a speed of \(4.00 \mathrm{~m} / \mathrm{s}\) ?

A \(3.50-\mathrm{kN}\) piano is lifted by three workers at constant speed to an apartment \(25.0 \mathrm{~m}\) above the street using a pulley system fastened to the roof of the building. Each worker is able to deliver 165 W of power, and the pulley system is \(75.0 \%\) efficient (so that \(25.0 \%\) of the mechanical energy is lost due to friction in the pulley). Neglecting the mass of the pulley, find the time required to lift the piano from the street to the apartment.

QIC A worker pushing a \(35.0-\mathrm{kg}\) wooden crate at a constant speed for \(12.0 \mathrm{~m}\) along a wood floor does \(350 \mathrm{~J}\) of work by applying a constant horizontal force of magnitude \(F_{0}\) on the crate. (a) Determine the value of \(F_{0}\) - (b) If the worker now applies a force greater than \(F_{0}\), describe the subsequent motion of the crate. (c) Describe what would happen to the crate if the applied force is less than \(F_{0}\).

GP A \(60.0-\mathrm{kg}\) athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0\). What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

When a \(2.50-\mathrm{kg}\) object is hung vertically on a certain light spring described by Hooke's law, the spring stretches \(2.76 \mathrm{~cm}\). (a) What is the force constant of the spring? (b) If the \(2.50-\mathrm{kg}\) object is removed, how far will the spring stretch if a \(1.25-\mathrm{kg}\) block is hung on it? (c) How much work must an external agent do to stretch the same spring \(8.00 \mathrm{~cm}\) from its unstretched position?

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