A rock with mass m=3.00 kg falls from rest in a viscous medium. The rock is acted on by a net co

step 1 Question number 101. For part A, we have to find the acceleration A -0. Excellation A -0 is equa

A rock with mass m=3.00 kg falls from rest in a viscous...

A rock with mass $m=3.00 \mathrm{kg}$ 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 $\mathrm{m} \mathrm{m} / \mathrm{s}$ and $k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}$ (see Section 5.3$) .$ (a) Find the initial acceleration $a_{0} \cdot$ (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 $v_{t}$ (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 0.9$v_{\mathrm{t}}$

A rock with mass $m=3.00 \mathrm{kg}$ 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 $\mathrm{m} \mathrm{m} / \mathrm{s}$ and $k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}$ (see Section 5.3$) .$ (a) Find the initial acceleration $a_{0} \cdot$ (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 $v_{t}$ (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 0.9$v_{\mathrm{t}}$

Key Concepts

Differential Equations in Motion

Motion under a combination of constant and velocity-dependent forces is typically modeled by a first-order linear differential equation (m dv/dt = F_net - k·v). This mathematical framework allows for the determination of velocity, acceleration, and displacement as functions of time, and is essential for solving problems involving objects moving through a viscous medium.

Kinematics Under Resistive Forces

When an object moves in a medium that exerts a velocity-dependent drag force, its acceleration is not constant but changes as the speed increases. Understanding the interplay between the constant driving forces and the variable resistive force involves using differential equations to describe how velocity and position change over time.

Viscous Drag Force

The drag force in a viscous medium is often modeled as being proportional to the speed of the object. In such cases, the force can be expressed as f = k·v, where k is a constant that characterizes the resistance of the medium. This concept is critical for understanding how resistive forces modify the motion of falling objects.

Newton's Second Law

Newton's Second Law establishes that the net force acting on an object is equal to its mass multiplied by its acceleration. This principle is used to relate the forces in motion problems—including gravitational force, buoyant force, and drag force—to the resultant acceleration of the object.

Terminal Velocity

Terminal velocity is reached when the force due to drag balances the net driving force (for example, gravity minus buoyancy), resulting in zero net acceleration. At this point, the object continues to fall at a constant speed. This concept is essential when analyzing motion under resistive forces because it represents the long-term behavior of the system.

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