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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 5, Problem 21c

When jumping straight up from a crouched position, an average person can reach a maximum height of about 6060 cm. During the jump, the person's body from the knees up typically rises a distance of around 5050 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. In terms of this jumper's weight w, what force does the ground exert on him or her during the jump?

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Step 1: Identify the forces acting on the jumper during the jump. The ground exerts an upward force (normal force) on the jumper, which must overcome the jumper's weight (gravitational force) and provide the additional force needed to accelerate the jumper upward.
Step 2: Use Newton's second law of motion, which states that the net force is equal to the mass times the acceleration: F=ma. Here, the net force is the difference between the upward force exerted by the ground and the jumper's weight.
Step 3: Calculate the acceleration of the jumper during the jump. Use the kinematic equation: v22=u22+2as, where v is the final velocity (0 m/s at the peak of the jump), u is the initial velocity, a is the acceleration, and s is the displacement (50 cm or 0.5 m). Rearrange the equation to solve for a.
Step 4: Relate the jumper's weight to their mass using the equation w=mg, where g is the acceleration due to gravity (approximately 9.8 m/s²). Substitute this into the net force equation to express the upward force exerted by the ground in terms of the jumper's weight.
Step 5: Combine the equations from Steps 2 and 4 to find the total force exerted by the ground. This force is equal to the jumper's weight plus the additional force required to accelerate the jumper upward. The final expression will be in terms of the jumper's weight w.

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Key Concepts

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

Newton's Third Law of Motion

Newton's Third Law states that for every action, there is an equal and opposite reaction. In the context of jumping, when a person exerts a force downward on the ground, the ground exerts an equal force upward on the person. This reaction force is crucial for understanding how the jumper is propelled into the air.
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Weight and Gravitational Force

Weight is the force exerted by gravity on an object, calculated as the product of mass and the acceleration due to gravity (w = mg). In this scenario, the jumper's weight plays a significant role in determining the net force acting on them during the jump, influencing how high they can ascend.
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Net Force and Acceleration

The net force acting on an object is the vector sum of all forces acting on it. According to Newton's second law (F = ma), this net force determines the object's acceleration. During the jump, the ground's upward force must overcome the jumper's weight to create a net upward force, allowing the jumper to accelerate upwards.
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Related Practice
Textbook Question

A 5.005.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t)F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=y(t) = (2.802.80 m/s)t + (0.6100.610 m/s3)t3. What is the magnitude of FF when t=4.00t = 4.00 s?

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Textbook Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about 6060 cm. During the jump, the person's body from the knees up typically rises a distance of around 5050 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. Draw a free-body diagram of the person during the jump.

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Textbook Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about 6060 cm. During the jump, the person's body from the knees up typically rises a distance of around 5050 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. With what initial speed does the person leave the ground to reach a height of 6060 cm?

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Textbook Question

In a laboratory experiment on friction, a 135135-N block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the block begins to move and continues to increase thereafter. Figure E5.265.26 shows a graph of the friction force on this block as a function of the pull. Identify the regions of the graph where static friction and kinetic friction occur.

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Textbook Question

A 2.002.00-kg box is moving to the right with speed 9.009.00 m/s on a horizontal, frictionless surface. At t=0t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)=F(t) = (6.006.00 N/s2)t2. What distance does the box move from its position at t=0t = 0 before its speed is reduced to zero?

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Textbook Question

A 550550-N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 850850 kg. As the elevator starts moving, the scale reads 450450 N. Find the acceleration of the elevator (magnitude and direction).

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