Textbook Question
A 2.0 kg object is moving to the right with a speed of when it experiences the force shown in FIGURE EX11.9. What are the object's speed and direction after the force ends?
Ch 11: Impulse and Momentum
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 11, Problem 5
In FIGURE EX11.5, what value of Fmax gives an impulse of 6.0 N s?
Verified step by step guidance1
Step 1: Recall the formula for impulse, which is given by \( J = F \cdot \Delta t \), where \( J \) is the impulse, \( F \) is the force, and \( \Delta t \) is the time interval during which the force acts.
Step 2: From the graph, observe that the force \( F_{\text{max}} \) acts over a time interval \( \Delta t \) from \( t = 4 \, \text{s} \) to \( t = 10 \, \text{s} \). Calculate \( \Delta t \) as \( \Delta t = 10 - 4 = 6 \, \text{s} \).
Step 3: Substitute the given impulse \( J = 6.0 \, \text{N} \cdot \text{s} \) and the time interval \( \Delta t = 6 \, \text{s} \) into the formula \( J = F \cdot \Delta t \). Rearrange the formula to solve for \( F \): \( F = \frac{J}{\Delta t} \).
Step 4: Perform the substitution: \( F = \frac{6.0}{6} \). This will give the value of \( F_{\text{max}} \).
Step 5: Interpret the result: The calculated \( F_{\text{max}} \) represents the maximum constant force that acts over the time interval \( \Delta t \) to produce the given impulse.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Impulse
Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It is mathematically represented as the product of force and the time duration for which the force acts. In this context, the impulse is given as 6.0 N·s, indicating the total effect of the force applied over the specified time interval.
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Force-Time Graph
A force-time graph visually represents how force varies with time. The area under the curve of this graph corresponds to the impulse experienced by an object. In the provided graph, the force is constant at Fmax for a certain duration, allowing for straightforward calculation of impulse by multiplying Fmax by the time interval during which the force is applied.
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Calculating Fmax
To find the maximum force (Fmax) that results in a specific impulse, one can use the relationship between impulse, force, and time. Given the impulse (6.0 N·s) and the duration of the force application (which can be inferred from the graph), Fmax can be calculated using the formula: Impulse = Fmax × time. This allows for determining the required force to achieve the desired impulse.
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Related Practice
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FIGURE EX11.6 is an incomplete momentum bar chart for a collision that lasts 10 ms. What are the magnitude and direction of the average collision force exerted on the object?
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What is the impulse on a 3.0 kg particle that experiences the force shown in FIGURE EX11.4?
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What impulse does the force shown in FIGURE EX11.3 exert on a 250 g particle?
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Textbook Question
A 2.0 kg object is moving to the right with a speed of 1.0 m/s when it experiences the force shown in FIGURE EX11.8. What are the object's speed and direction after the force ends?
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