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Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 07: Newton's Third LawProblem 12b
Chapter 7, Problem 12b

The foot of a 55 kg sprinter is on the ground for 0.25 s while her body accelerates from rest to 2.0 m/s. What is the magnitude of the friction force?

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Identify the known quantities: The mass of the sprinter is \( m = 55 \; \text{kg} \), the time during which the foot is on the ground is \( t = 0.25 \; \text{s} \), and the final velocity is \( v_f = 2.0 \; \text{m/s} \). The initial velocity is \( v_i = 0 \; \text{m/s} \).
Calculate the acceleration of the sprinter using the kinematic equation \( a = \frac{v_f - v_i}{t} \). Substituting the known values, \( a = \frac{2.0 - 0}{0.25} \).
Determine the net force acting on the sprinter using Newton's second law, \( F = m \cdot a \). Substituting the values for \( m \) and \( a \), \( F = 55 \cdot a \).
Recognize that the net force in this case is provided by the friction force between the sprinter's foot and the ground. Therefore, the magnitude of the friction force is equal to the net force calculated in the previous step.
Substitute the value of \( a \) from step 2 into the equation for \( F \) to find the magnitude of the friction force. Ensure the units are consistent and the result is expressed in newtons (N).

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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. In this scenario, understanding this law is crucial to calculate the net force acting on the sprinter as she accelerates.
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Friction Force

Friction force is the resistance that one surface or object encounters when moving over another. It is influenced by the nature of the surfaces in contact and the normal force acting between them. In this problem, the friction force is essential for providing the necessary acceleration to the sprinter, allowing her to reach a speed of 2.0 m/s.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. In this case, the kinematic equation can be used to determine the acceleration of the sprinter, which is necessary to find the net force and subsequently the friction force acting on her.
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Related Practice
Textbook Question

A 1000 kg car is pushing an out-of-gear 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N. Rolling friction can be neglected. What is the magnitude of the force of the car on the truck?

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Textbook Question

A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s2. How much force pulls forward on he rope? Assume that the rope is perfectly horizontal.

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Textbook Question

How much force does the astronaut exert on his chair while accelerating straight up at 10 m/s2?

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Textbook Question

The sled dog in FIGURE EX7.15 drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N, what is the tension in rope 2?

Textbook Question

Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. How much force does the 2 kg block exert on the 1 kg block?

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Textbook Question

FIGURE EX7.17 shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s2 by force F. What is the tension at the top end of rope 1?

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