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Chapter 3: Problem 52

A horizontal cable pulls a \(200-\mathrm{kg}\) cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, (a) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?

Short Answer

Expert verified
(a) 3.2 s; (b) 12.8 m

Step by step solution

01

Identify known values

We know the mass of the cart is \( m = 200 \, \text{kg} \), the tension (force) in the cable is \( F = 500 \, \text{N} \), and the final velocity \( v_f = 8.0 \, \text{m/s} \). The cart starts from rest, so its initial velocity \( v_i = 0 \).
02

Calculate acceleration using Newton's Second Law

Newton's second law states \( F = ma \), where \( a \) is the acceleration. We can rearrange this to find \( a \): \[ a = \frac{F}{m} = \frac{500 \, \text{N}}{200 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]
03

Use kinematic equation to find time

The formula for acceleration in terms of velocity and time is \( a = \frac{v_f - v_i}{t} \). We can solve for time \( t \): \[ t = \frac{v_f - v_i}{a} = \frac{8.0 \, \text{m/s} - 0}{2.5 \, \text{m/s}^2} = 3.2 \, \text{s} \]
04

Use a second kinematic equation to find displacement

To find the displacement \( s \), we use the equation \( s = v_i t + \frac{1}{2}at^2 \). Since \( v_i = 0 \), \[ s = \frac{1}{2} \times 2.5 \, \text{m/s}^2 \times (3.2 \, \text{s})^2 = 12.8 \, \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Kinematic equations are essential in physics for describing motion with constant acceleration. These equations connect variables such as displacement, initial and final velocity, acceleration, and time, making it possible to solve motion problems even when not all parameters are directly known.

There are four main kinematic equations. For this exercise, the relevant equations used are:
  • The formula connecting acceleration, initial and final velocities, and time: \[ a = \frac{v_f - v_i}{t} \]
  • The displacement equation when starting from rest: \[ s = \frac{1}{2}at^2 \]
By mastering these equations, you can tackle a wide range of motion-related problems. Applying the appropriate formula depending on the given and required quantities is key.

Remember, the kinematic equations assume constant acceleration, so ensure this condition is met in your problems.
Acceleration Calculation
The acceleration is a measure of how quickly an object changes its velocity. It is calculated using Newton's Second Law of Motion. This law states that the force applied to an object is equal to the mass of the object times its acceleration, expressed as \( F = ma \).

For our scenario, the tension in the cable (the force) is known, as well as the mass of the cart. Rearranging the formula gives the acceleration:
  • \[ a = \frac{F}{m} = \frac{500 \, \text{N}}{200 \, \text{kg}} = 2.5 \, \text{m/s}^2 \]
This calculation shows that every second, the cart's speed increases by 2.5 meters per second. Acceleration is a vector, meaning it has both a magnitude and a direction, and in this case, the direction is horizontal along the track.

Understanding how to calculate acceleration is critical when analyzing motion, as it gives insight into how quickly an object is changing its speed over time.
Displacement Calculation
Displacement is the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. For objects under constant acceleration starting from rest, one of the kinematic equations can directly compute displacement:

\[ s = v_i t + \frac{1}{2}at^2 \]

In this particular exercise, the initial velocity \( v_i \) is zero, simplifying the equation to:
  • \[ s = \frac{1}{2} \times 2.5 \, \text{m/s}^2 \times (3.2 \, \text{s})^2 = 12.8 \, \text{m} \]
This result tells us that in the 3.2 seconds it takes for the cart to reach a speed of 8 m/s, it travels 12.8 meters along the track.

Whenever you deal with problems involving displacement, consider both the initial conditions (such as starting from rest) and any external forces (like tension) that may be influencing the system's motion.

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Most popular questions from this chapter

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a \(900-\mathrm{g}\) mass. ( \(a\) ) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. (b) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

The mythical planet Mongo has twice the mass and twice the radius of Earth. Compute the acceleration due to gravity at its surface. We know from Problem \(3.40\) that in general $$ g_{R}=\frac{G M}{R^{2}} $$ Then for the Earth at its surface $$ g_{E}=\frac{G M_{E}}{R_{E}^{2}}=g=9.81 \mathrm{~m} / \mathrm{s}^{2} $$ where \(R_{E}\) is the Earth's radius and \(M_{E}\) is its mass. For Mongo $$ g_{M}=\frac{G M_{M}}{R_{M}^{2}}=\frac{G\left(2 M_{E}\right)}{\left(2 R_{E}\right)^{2}} $$ and or $$ \begin{array}{c} g_{M}=\frac{1}{2} \frac{G M_{E}}{R_{E}^{2}}=\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \\ g_{M}=4.91 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$

How large a horizontal force in addition to \(F_{T}\) must pull on block- \(A\) in Fig. \(3-21\) to give it an acceleration of \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) toward the left? Assume, as in Problem \(3.31\), that \(\mu_{k}=0.20, m_{A}=25 \mathrm{~kg}\), and \(m_{B}=15 \mathrm{~kg}\). Redraw Fig 3-21 for this case, including a force \(F\) pulling toward the left on \(A\). In addition, the retarding friction force \(F_{\mathrm{f}}\) must be reversed in direction. As in Problem 3.31, \(F_{\mathrm{f}}=49.1 \mathrm{~N}\). Write \(F=m a\) for each block in turn, taking the direction of motion (to the left and up) to be positive. We have $$ \begin{aligned} \pm \sum F_{x A}=F-F_{T}-49.1 \mathrm{~N} &=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \text { and }+\uparrow \sum F_{y B}=F_{T}-(15)(9.81) \mathrm{N} \\ &=(15 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ Solve the last equation for \(F_{T}\) and substitute in the previous equation. Then solve for the single unknown \(F\), and find it to be \(226 \mathrm{~N}\) or \(0.23 \mathrm{kN}\).

Three forces that act on a particle are given by \(\overrightarrow{\mathbf{F}}_{1}=(20 \hat{\mathbf{i}}-36 \hat{\mathbf{j}}+73 \hat{\mathbf{k}}) \mathbf{N}, \overrightarrow{\mathbf{F}}_{2}=(-17 \hat{\mathbf{i}}+21 \hat{\mathbf{j}}-46 \hat{\mathbf{k}})\) \(\mathrm{N}\), and \(\overrightarrow{\mathbf{F}}_{3}=(-12 \hat{\mathbf{k}}) \mathrm{N}\). Find their resultant vector. Also find the magnitude of the resultant to two significant figures.We know that $$ \begin{array}{l} R_{x}=\sum F_{x}=20 \mathrm{~N}-17 \mathrm{~N}+0 \mathrm{~N}=3 \mathrm{~N} \\ R_{y}=\sum F_{y}=-36 \mathrm{~N}+21 \mathrm{~N}+0 \mathrm{~N}=-15 \mathrm{~N} \\\ R_{z}=\sum F_{z}=73 \mathrm{~N}-46 \mathrm{~N}-12 \mathrm{~N}=15 \mathrm{~N} \end{array} $$ Since \(\overrightarrow{\mathbf{R}}=R_{x} \hat{\mathbf{i}}+R_{y} \hat{\mathbf{j}}+R_{z} \hat{\mathbf{k}}\), $$ \overrightarrow{\mathbf{R}}=3 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}} $$ To two significant figures, the three-dimensional Pythagorean theorem then gives $$ R=\sqrt{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}=\sqrt{459}=21 \mathrm{~N} $$

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).
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