A cart that can move along a straight track has a mass of 3.00 kg and an initial velocity of 4.00 m/s in the negative x-direction. The cart is then subjected to a net force that is initially in the positive x-direction, but which then eventually switches direction to the negative x-direction, as shown in the graph of force as a function of time below. The force is directed only in the +x or -x direction. F, (N)
A cart that can move along a straight track has a mass of 3.00 kg and an initial velocity of 4.00 m/s in the negative x-direction. The cart is then subjected to a net force that is initially in the positive x-direction, but which then eventually switches direction to the negative x-direction, as shown in the graph of force as a function of time below. The force is directed only in the +x or -x direction. F, (N)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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
Transcribed Image Text:**Title: Understanding Force and Momentum on a Moving Cart**
**Introduction:**
A cart with a mass of 3.00 kg and an initial velocity of 4.00 m/s in the negative x-direction is subjected to various net forces over time. The forces alternate between positive and negative directions, influencing the cart's momentum and velocity.
**Graph Explanation:**
The graph illustrates the force \( F_x \) (in Newtons) applied to the cart as a function of time \( t \) (in seconds). The x-axis represents time from 0 to 30 seconds, while the y-axis represents the force from -20 N to +30 N. Key sections include:
- From 0 to approximately 7 seconds, the force decreases from +10 N to 0 N.
- From 7 to 12 seconds, the force remains constant at 0 N.
- At 12 seconds, there is a sudden change to -20 N, which remains steady until approximately 27 seconds, where it returns to 0 N.
**Table of Momentum and Velocity:**
| Time | Momentum | Velocity |
|----------|-----------------|----------------|
| \( t = 0 \, \text{s} \) | \( p = -12 \, \text{kg} \cdot \text{m/s} \) ✔️ | \( v = -4 \, \text{m/s} \) ✔️ |
| \( t = 20 \, \text{s} \) | \( p = 88 \, \text{kg} \cdot \text{m/s} \) | \( v = 22 \, \text{m/s} \) ❌ |
| \( t = 25 \, \text{s} \) | \( p = 38 \, \text{kg} \cdot \text{m/s} \) | \( v = 9.5 \, \text{m/s} \) |
**Question:**
(b) The cart's momentum at \( t = 20 \, \text{s} \) is exactly the same as its momentum at \( t = \_\_\_\_ \, \text{s} \).
**Conclusion:**
This example illustrates how forces acting in different directions over time influence both the momentum and velocity of an object. Understanding these changes is crucial in studying motion dynamics and applying Newton’s laws effectively.
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