Problem 108 - University Physics with Modern Physics

Question

A rock with mass \(m\) slides with initial velocity \(v_{0}\) on a horizontal surface. A retarding force \(F_{\mathrm{R}}\) that the surface exerts on the rock is proportional to the square root of the instantaneous velocity of the rock \(\left(F_{\mathrm{R}}=-k v^{1 / 2}\right)\) . (a) Find expressions for the velocity and position of the rock as a function of time. (b) In terms of \(m, k,\) and \(v_{0},\) at what time will the rock come to rest? (c) In terms of \(m, k,\) and \(v_{0},\) what is the distance of the rock from its starting point when it comes to rest?

Step-by-Step Solution

Answer

(a) \(v(t) = \left(\frac{2m \cdot v_0^{1/2} - kt}{2m}\right)^2\), (b) \(t = \frac{2m \cdot v_0^{1/2}}{k}\), (c) \(x = \frac{4m^2}{3k^2} \cdot v_0^{3/2}\)."

Step 1: Set up Newton’s Second Law

Newton's second law states that the net force acting on an object is equal to the mass of the object times its acceleration, or \(F = ma\). Here, the retarding force \(F_{\mathrm{R}}\) can be expressed as \(m \cdot \frac{dv}{dt} = -k v^{1/2}\).

Step 2: Separate and integrate for velocity

Rearrange the differential equation to separate the variables: \(m \cdot dv = -k v^{1/2} \cdot dt\). Now, integrate both sides. The left side becomes \(\int m \cdot v^{-1/2} \cdot dv\), and the right side becomes \(\int -k \cdot dt\).

Step 3: Solve the integral for velocity

The integral of the left side is \(2m \cdot v^{1/2}\), and of the right side is \(-kt + C_1\). Equating them gives \(2m \cdot v^{1/2} = -kt + C_1\). At \(t = 0\), \(v = v_0\), therefore \(C_1 = 2m \cdot v_0^{1/2}\). Substituting back gives \(v = \left(\frac{C_1 - kt}{2m}\right)^2\).

Step 4: Find the expression for position

Start with \(v = \frac{dx}{dt}\) and replace the expression for \(v\) from Step 3. Solve \(\frac{dx}{dt} = \left(\frac{2m \cdot v_0^{1/2} - kt}{2m}\right)^2\) to find \(x\). Integrate both sides with respect to \(t\): \(x = \int \frac{1}{4m^2} \cdot (2m \cdot v_0^{1/2} - kt)^2\, dt\).

Step 5: Determine when the rock comes to rest

The rock comes to rest when \(v = 0\). From the velocity expression \(v = \left(\frac{2m \cdot v_0^{1/2} - kt}{2m}\right)^2\), set it to 0 and solve for \(t\): \(kt = 2m \cdot v_0^{1/2}\). Thus, \(t = \frac{2m \cdot v_0^{1/2}}{k}\).

Step 6: Calculate the distance when rock comes to rest

Substitute \(t = \frac{2m \cdot v_0^{1/2}}{k}\) into the position equation derived in Step 4 to find \(x\) at rest. Evaluate the integral, finding that the distance \(x = \frac{4m^2}{3k^2} \cdot v_0^{3/2}\).

Key Concepts

Retarding Force

When studying motion, a retarding force is any force that opposes the direction of motion of an object, effectively slowing it down. In the context of the rock sliding on a surface, this retarding force is unique because it is proportional to the square root of the rock's velocity. This means the force isn't constant; it changes depending on how fast the rock is moving. The force is described by the equation

• \( F_{\mathrm{R}} = -k v^{1/2} \)

where \( k \) is a constant that scales the force's strength, and \( v \) is the instantaneous velocity of the rock. The negative sign indicates that the force is acting in the opposite direction of the velocity, thus slowing the rock down.
Retarding forces like this appear in real-life scenarios such as frictional forces or drag forces in fluids, where the nature of the opposing force depends on the speed of the moving object.

Differential Equations

Differential equations are mathematical tools used to describe relationships involving rates of change. In this exercise, the retarding force creates an equation that involves the rate of change of the rock's velocity. By Newton's Second Law, the force on the rock is described by

• \( m \cdot \frac{dv}{dt} = -k v^{1/2} \)

This is a differential equation because it relates the velocity \( v \) and its rate of change with respect to time \( \frac{dv}{dt} \).
The equation must be solved to find how the velocity of the rock changes over time. Solving it involves separating the variables, allowing each side of the equation to be integrated, which will yield a function describing how the velocity changes as time progresses.

Velocity as a Function of Time

Velocity as a function of time tells us how the speed of the rock evolves after it starts sliding. Solving the differential equation reveals that the velocity \( v \) is given by

• \( v = \left( \frac{2m \cdot v_0^{1/2} - kt}{2m} \right)^2 \)

This expression shows how the initial velocity \( v_0 \) and the retarding force impact the speed over time. Initially, time \( t \) is zero, so the expression reduces to the initial conditions. As time increases, the term \( -kt \) diminishes the initial term, indicating that the velocity decreases until the rock comes to rest.
Understanding this function is crucial for predicting motion in physics, allowing us to determine how long the object will take to stop or how far it will travel.

Integration in Physics

Integration is a fundamental concept in physics used to sum up incremental changes to find quantities like position when the rate of change (like velocity) is known. After deriving the velocity function, we can determine the rock's position over time by solving the integral of its velocity. The position \( x \) function is derived from evaluating

• \( x = \int \left( \frac{2m \cdot v_0^{1/2} - kt}{2m} \right)^2 \, dt \)

This integration yields the total distance traveled by the rock over time.
In the given exercise, this integration helps determine how far the rock will have moved by the time it comes to rest. Such calculations are pivotal in physics, providing insights into how different forces and initial conditions affect motion. Through integration, one can bridge the gap between instantaneous velocities and actual pathway distances in the physical world.