Problem Find the particle's horizontal position x(t) and velocity v(x) at any point in a fluid whose drag force is expressed as Fdrag = kmv where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F--F -m a Then: -m V=m a since: a = dv/dt then -m v=m dv/dt by integrating, we obtain the following expression: = voe Further, employing the rules of integration results to the following expression for position as a function of time x= (vo/ as t→ ∞, the position becomes x = vo/k b) To solve for the velocity as a function of position v(x), we construct the net force in the x-axis as follows E F--F Then: -m V = m

Problem Find the particle's horizontal position x(t) and velocity v(x) at any point in a fluid whose drag force is expressed as Fdrag = kmv where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F--F -m a Then: -m V=m a since: a = dv/dt then -m v=m dv/dt by integrating, we obtain the following expression: = voe Further, employing the rules of integration results to the following expression for position as a function of time x= (vo/ as t→ ∞, the position becomes x = vo/k b) To solve for the velocity as a function of position v(x), we construct the net force in the x-axis as follows E F--F Then: -m V = m

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter6: Applications Of Newton’s Laws Of Motion

Section: Chapter Questions

Problem 62PQ

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Problem Find the particle's horizontal position x(t) and velocity v(x) at any point in a fluid whose drag force is expressed as Fdrag = kmv where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity v0. To solve for the position as a function of time x(t), construct the net force in the x-axis. Further, employ the rules of integration results to the following expression for position as a function of time. To solve for the velocity as a function of position v(x), we construct the net force in the x-axis, then, eliminate time by expressing, the velocity on the left side of the equation. By integrating and applying the limits, we then shall arrive with a velocity that decreases in a linear maner.
Problem Find the particle's horizontal position x(t) and velocity v(x) at any point in a fluid whose drag force is expressed as Fdrag = kmv where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity v0. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as ∑F = -F_____ = m_____ Then: -m_____v = m_____ since: a = dv/dt then -m_____v = m_____ by integrating, we obtain the following expression: _____ = v0e_____ Further, employing the rules of integration results to the following expression for position as a function of time x= (v0/_____)( _____ - e _____ ) as t -> ∞ , the position becomes x = v0/k b) To solve for the velocity as a function of position v(x), we construct the net force in the x-axis as follows ∑F = -F _____= m_____ Then: -m_____v = m_____ since: a = dv/dt then -m_____v = m_____ We can eliminate time by expressing, the velocity…
Like friction, drag force opposes the motion of a particle in a fluid; however, drag force depends on the particle's velocity. Find the expression for the particle's velocity v(x) as a function of position at any point x in a fluid whose drag force is expressed as Fdrag = kmv where k is a constant, m is the mass of the particle and v is its velocity. Assume that the particle is constrained to move in the x-axis only with an initial velocity v0. Solution: The net force along the x-axis is: ΣF = -F = m then: -mv = m Since acceleration is the first time derivative of velocity a = dv/dt, -mv = m We can eliminate time dt by expressing, the velocity on the left side of the equation as v = dx/dt. Manipulating the variables and simplifying, we arrive at the following expression / = -k "Isolating" the infinitesimal velocity dx and integrating with respect to dx, we arrive at the following: = v0 - which shows that velocity decreases in a linear manner.
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