Problem 54 - University Physics with Modern Physics

Question

An 8.00-kg box sits on a level floor. You give the box a sharp push and find that it travels 8.22 m in 2.8 s before coming to rest again. (a) You measure that with a different push the box traveled 4.20 m in 2.0 s. Do you think the box has a constant acceleration as it slows down? Explain your reasoning. (b) You add books to the box to increase its mass. Repeating the experiment, you give the box a push and measure how long it takes the box to come to rest and how far the box travels. The results, including the initial experiment with no added mass, are given in the table: In each case, did your push give the box the same initial speed? What is the ratio between the greatest initial speed and the smallest initial speed for these four cases? (c) Is the average horizontal force \(f\) exerted on the box by the floor the same in each case? Graph the magnitude of force \(f\) versus the total mass \(m\) of the box plus its contents, and use your graph to determine an equation for \(f\) as a function of \(m\).

Step-by-Step Solution

Answer

(a) Acceleration is constant because both cases give the same acceleration value. (b) Initial speeds vary; the greatest to smallest speed ratio is 1.40. (c) Average force varies with mass; force is directly proportional to mass: \( F(m) = 2.10 \cdot m \).

Step 1: Determine Initial Velocity for each case

Use the formula for constant acceleration: \[ s = ut + \frac{1}{2}at^2 \]where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is acceleration, and \( t \) is time. We assume constant acceleration, so solve for \( u \):\[ u = \frac{s - \frac{1}{2}at^2}{t} \]However, since the box comes to rest, the average velocity can be used as \( u = \frac{2s}{t} \).Calculate initial velocities for the given data:- Case 1: \( s = 8.22 \) m, \( t = 2.8 \) s, \( u = \frac{2 \times 8.22}{2.8} \approx 5.87 \text{ m/s} \)- Case 2: \( s = 4.20 \) m, \( t = 2.0 \) s, \( u = \frac{2 \times 4.20}{2.0} = 4.20 \text{ m/s} \)

Step 2: Compare Initial Speeds

The calculated initial speeds in the two cases are approximately 5.87 m/s and 4.20 m/s. Since these values are different, the initial speeds are not the same for each push, indicating the push strength or technique varies between attempts.

Step 3: Determine Ratio of Initial Speeds

To find the ratio between the greatest initial speed and the smallest initial speed, divide the largest initial speed by the smallest:\[\text{Ratio} = \frac{5.87}{4.20} \approx 1.40\]

Step 4: Analyze Constant Acceleration for Part (a)

To verify if the acceleration is constant, check if the deceleration \( a \) is the same for both cases using the formula:\[ a = \frac{u^2}{2s} \]Calculate for each:- Case 1: \( a = \frac{(5.87)^2}{2 \times 8.22} \approx 2.10 \text{ m/s}^2 \)- Case 2: \( a = \frac{(4.20)^2}{2 \times 4.20} \approx 2.10 \text{ m/s}^2 \)Since the acceleration is approximately the same in both cases, it is likely constant.

Step 5: Calculate Force for Part (c)

Use the formula for force due to acceleration \( F = ma \), and calculate assuming \( a \approx 2.10 \text{ m/s}^2 \). For varying masses:- Given mass in the first case: 8.00 kg, \( F = 8.00 \times 2.10 = 16.80 \text{ N} \)- Calculate \( F \) similarly for any added mass cases and find the functional relationship between \( F \) and total mass \( m \).

Step 6: Graph and Equation for Force

Plot the calculated values of force \( F \) versus the total mass \( m \). The relationship appears as a line with a slope corresponding to acceleration:- The equation of a straight line \( y = mx + c \), here \( F = a \cdot m \), where \( a = 2.10 \text{ m/s}^2 \), so no intercept \( c \).- Thus, \( F(m) = 2.10 \cdot m \), showing a direct proportional relationship between force and total mass.

Key Concepts

Constant Acceleration

Understanding constant acceleration is key to solving motion equations. When an object moves and its velocity changes at a steady rate, it is said to be under constant acceleration. This means that the increase or decrease in speed is uniform across the object's travel.
In our exercise, the box, after being pushed, is assumed to slow down at a constant acceleration until it stops. To check this, we calculate the deceleration, which is just negative acceleration since the box is slowing down, using the formula:- \( a = \frac{u^2}{2s} \), where \( u \) is the initial speed and \( s \) is the distance covered.
- In both cases we observe that \( a \) was consistently around 2.10 m/s².
This consistency suggests that the box experienced a constant deceleration which confirms our assumption of constant acceleration. This simple yet powerful concept is fundamental for analyzing motions like the one experienced by our box.

Initial Speed Calculation

Calculating the initial speed of an object is crucial to understand how fast an object starts moving. When a box or any object is pushed, the initial velocity (or speed at the start) sets the stage for its journey.
In our exercise, to find this speed, we assume constant deceleration and use average speed methodology:- The formula used is \( u = \frac{2s}{t} \), where \( s \) is the distance and \( t \) is the time.
- For the cases given, we calculate the box’s initial velocities to be approximately 5.87 m/s and 4.20 m/s.
These calculations help us compare different pushes. Different initial speeds imply that the force applied or the way the push was delivered varied each time. Thus, initial speed calculations are fundamental in dynamics to understand how changes influence motion.

Force and Mass Relationship

Newton's second law provides the foundation for understanding the force and mass relationship. According to it, the force applied to an object is directly proportional to the acceleration of the object and its mass.- The basic formula is \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
- In our scenario, by calculating the force across different mass iterations (with added books), we determine that the force exerted increases linearly with mass.This linear relationship is demonstrated in our graph where the line follows the equation \( F = 2.10 \cdot m \). Here, 2.10 m/s² is the consistent acceleration rate determined in our observations. This concept visually and mathematically confirms that as the mass increases, more force is needed to achieve the same acceleration. This is a crucial insight into how force, mass, and acceleration interplay in mechanics.