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Ch 04: Newton's Laws of Motion
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 4, Problem 13b

A 4.504.50-kg experimental cart undergoes an acceleration in a straight line (the xx-axis). The graph in Fig. E4.134.13 shows this acceleration as a function of time. During what times is the net force on the cart a constant?
Graph depicting acceleration as a function of time, relevant to Newton's laws.

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Step 1: Analyze the graph provided. The acceleration (a) is plotted as a function of time (t). The graph shows regions where the acceleration is constant, which corresponds to a constant net force on the cart. According to Newton's second law, the net force (F) is given by F = m * a, where m is the mass of the cart and a is the acceleration.
Step 2: Identify the regions of constant acceleration from the graph. In the first graph, the acceleration is constant between 2 ms and 6 ms (a = 2 m/s²). In the second graph, the acceleration is constant between 2 s and 4 s (a = 10 m/s²). These intervals indicate that the net force is constant during these times.
Step 3: Recall that when acceleration changes (either increasing or decreasing), the net force is not constant. For example, in the first graph, the acceleration decreases from 6 m/s² to 2 m/s² between 0 ms and 2 ms, and decreases again from 2 m/s² to 0 m/s² between 8 ms and 12 ms. Similarly, in the second graph, the acceleration increases and decreases outside the interval of constant acceleration.
Step 4: Use Newton's second law to confirm the relationship between force and acceleration. Since the mass of the cart is constant (4.50 kg), the net force is directly proportional to the acceleration. Therefore, during the intervals of constant acceleration, the net force remains constant.
Step 5: Summarize the findings. The net force on the cart is constant during the intervals where the acceleration is constant: 2 ms to 6 ms in the first graph, and 2 s to 4 s in the second graph. Outside these intervals, the net force varies due to changes in acceleration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. Understanding this law is crucial for analyzing the forces acting on the experimental cart as it accelerates.
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Intro to Forces & Newton's Second Law

Acceleration

Acceleration is defined as the rate of change of velocity of an object with respect to time. It can be constant or variable, and in this context, it is represented as a function of time in the provided graph. The graph shows how acceleration changes, which helps determine when the net force on the cart remains constant, as constant acceleration indicates a constant net force.
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Graph Interpretation

Interpreting graphs is essential in physics for understanding relationships between variables. In this case, the graph depicts acceleration (y-axis) as a function of time (x-axis). By analyzing the graph, one can identify intervals where the acceleration is constant, which corresponds to periods where the net force acting on the cart is also constant, as per Newton's Second Law.
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Related Practice
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Textbook Question

A 4.504.50-kg experimental cart undergoes an acceleration in a straight line (the xx-axis). The graph in Fig. E4.134.13 shows this acceleration as a function of time. Find the maximum net force on this cart. When does this maximum force occur?

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Textbook Question

A hockey puck with mass 0.1600.160 kg is at rest at the origin (x=0x = 0) on the horizontal, frictionless surface of the rink. At time t=0t = 0 a player applies a force of 0.2500.250 N to the puck, parallel to the xx-axis; she continues to apply this force until t=2.00t = 2.00 s. What are the position and speed of the puck at t=2.00t = 2.00 s?

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