Problem 107 - University Physics with Modern Physics

Question

A rock with mass \(m=3.00\) kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v\) where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}(\) see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0 .}\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{\mathrm{t}}\) . (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed of 0.9\(v_{t} .\)

Step-by-Step Solution

Answer

The initial acceleration is 6.00 m/s². The terminal speed is 8.18 m/s, and the time to reach 0.9 times terminal speed is approximately 2.76 seconds.

Step 1: Determine Initial Acceleration

To find the initial acceleration, use Newton's second law. The net force is given as 18.0 N. At the start, velocity \(v = 0\), so the drag force \(f = 0\). Therefore, the acceleration \(a_0\) can be calculated by using the formula \(F_{net} = ma\): \[ a_0 = \frac{F_{net}}{m} = \frac{18.0 \text{ N}}{3.00 \text{ kg}} = 6.00 \text{ m/s}^2 \]

Step 2: Find Acceleration at Speed 3.00 m/s

To determine the acceleration when the speed is 3.00 m/s, calculate the fluid resistance force: \[ f = k v = 2.20 \text{ N·s/m} \times 3.00 \text{ m/s} = 6.60 \text{ N} \] The net force (after considering the fluid resistance) is: \[ F_{net} = 18.0 \text{ N} - 6.60 \text{ N} = 11.4 \text{ N} \] Now calculate the acceleration:\[ a = \frac{F_{net}}{m} = \frac{11.4 \text{ N}}{3.00 \text{ kg}} = 3.80 \text{ m/s}^2 \]

Step 3: Find Speed When Acceleration Equals 0.1a0

First calculate \( 0.1a_0 = 0.1 \times 6.00 \text{ m/s}^2 = 0.60 \text{ m/s}^2 \). The net force should be \( F_{net} = m \times 0.60 \text{ m/s}^2 = 1.80 \text{ N} \). Thus, the fluid resistance force when acceleration equals \( 0.1a_0 \) is:\[ f = 18.0 \text{ N} - 1.80 \text{ N} = 16.2 \text{ N} \] Calculate the speed:\[ k v = 16.2 \text{ N} \] \[ v = \frac{16.2 \text{ N}}{2.20 \text{ N·s/m}} = 7.36 \text{ m/s} \]

Step 4: Find Terminal Speed

At terminal speed, the acceleration is zero, so the resistance force equals the net force. \[ k v_t = 18.0 \text{ N} \] \[ v_t = \frac{18.0 \text{ N}}{2.20 \text{ N·s/m}} = 8.18 \text{ m/s} \]

Step 5: Compute Motion Parameters After 2 Seconds

Use the equation \( F_{net} = m \times a = m \times \frac{dv}{dt} \) with separation of variables to integrate and find the velocity function for time \( t \). The solution for velocity as a function of time, given the conditions, is: \[ v(t) = v_t(1 - e^{-\frac{k}{m}t}) \]Substitute \( t = 2 \), \( v_t = 8.18 \text{ m/s} \), \(k = 2.20 \text{ N·s/m}\), and \(m = 3.00 \text{ kg}\) to find: \[ v(2) = 8.18(1 - e^{-\frac{2.20}{3.00} \cdot 2}) = 6.85 \text{ m/s} \]Substitute back to find acceleration using: \[ a(t) = \frac{F_{net} - kv}{m} = \frac{18.0 - 2.20v}{3.00} \]

Step 6: Find Time to Reach 0.9 Terminal Speed

Set \(v = 0.9v_t\), which results in \(v = 0.9 \times 8.18\). Use the velocity equation:\[ 0.9 \times 8.18 = 8.18(1 - e^{-\frac{k}{m}t}) \]Solving for \(t\):\[ 0.9 = 1 - e^{-\frac{2.20}{3.00}t} \]\[ e^{-\frac{2.20}{3.00}t} = 0.1 \]\[ -\frac{2.20}{3.00}t = \ln(0.1) \]\[ t = -\frac{3.00}{2.20} \ln(0.1) \approx 2.76 \text{ seconds} \]

Key Concepts

Newton's Second Law

Newton's Second Law is a foundational principle in physics that explains how the force acting on an object is related to its mass and the acceleration it experiences. The law is succinctly expressed with the formula:\[ F_{net} = ma \]where:- \( F_{net} \) is the net force applied to the object,- \( m \) is the mass of the object,- \( a \) is the acceleration.In this scenario, with a falling rock in a fluid, the net force is the combination of gravitational and buoyant forces. Initially, the fluid resistance is zero because the rock starts from rest. Thus, each part of the problem revolves around applying this law, often adjusting for changing resistance as speed increases.

Fluid Resistance

Fluid resistance, or drag force, is an essential concept in understanding motion through a fluid, such as air or water. The rock experiences a resistance force given by the formula:\[ f = kv \]where:- \( f \) is the fluid resistance force,- \( k \) is a constant that represents characteristics of the medium, such as viscosity,- \( v \) is the velocity of the object.As the rock increases speed, the resistance force increases linearly with velocity. It acts in the opposite direction to the motion, reducing the net force and thus the acceleration. Calculating this force at different speeds allows us to understand how quickly and effectively the rock is slowed by the fluid.

Terminal Velocity

Terminal velocity occurs when an object falling through a fluid reaches such a high speed that the drag force equals the net force downwards, resulting in zero acceleration. At this point, the object continues to move at a constant speed. For the rock:The balance can be mathematically described as:\[ kv_t = F_{net} \]Substituting the values in for this exercise, we find the terminal velocity \( v_t \) by rearranging:\[ v_t = \frac{F_{net}}{k} \]Here, the use of terminal velocity is crucial to determine how the velocity changes over time and to find at what time the rock reaches 90% of this velocity.

Acceleration Calculation

Calculating acceleration in this context involves understanding both constant and variable forces acting on an object. Initially, the acceleration is simply the net force divided by the rock’s mass. But as speed increases, fluid resistance requires that we adjust this calculation:\[ a = \frac{F_{net} - kv}{m} \]where:- \( F_{net} \) is the constant net force downward,- \( kv \) is the fluid resistance,- \( m \) is the mass.By knowing the speed of the rock at any point, this formula helps compute the new acceleration. It lets us predict changes in speed over time, revealing how quickly the rock approaches its terminal velocity, or how quickly it changes speed based on time elapsed or resistance encountered.