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Ch 05: Applying Newton's Laws
Chapter 5, Problem 5

When jumping straight up from a crouched position, an average person can reach a maximum height of about 60 cm. During the jump, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (c) 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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Identify the initial and final velocities: When jumping, the initial velocity is what you need to find, and the final velocity at the maximum height is 0 m/s because the jumper momentarily stops moving upward.
Use the kinematic equation to relate the initial velocity, final velocity, acceleration due to gravity, and the height: \(v_f^2 = v_i^2 + 2a imes d\), where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration (gravity, which is -9.8 m/s^2), and \(d\) is the distance risen (0.5 m).
Solve the kinematic equation for the initial velocity \(v_i\): \(v_i = \\sqrt{v_f^2 - 2ad}\).
Calculate the force exerted by the ground using Newton's second law, which states that the force is equal to mass times acceleration: \(F = m imes a\). Here, the acceleration is not just gravity but the net acceleration during the jump.
Determine the net acceleration by considering the change in velocity over time. The time can be estimated using the motion equations, and then use \(F = m imes a\) to find the force, where \(m\) is the mass of the jumper (which can be found using \(w = m imes g\), where \(w\) is the weight and \(g\) is the acceleration due to gravity).

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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 force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). In the context of the jump, this law helps us understand how the force exerted by the ground must counteract both the weight of the jumper and provide the necessary upward acceleration to achieve the jump height.
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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). For the jumper, this weight acts downward, and understanding this force is crucial for determining the net force exerted by the ground during the jump, as it must overcome this weight to initiate upward motion.
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Net Force and Acceleration

The net force is the total force acting on an object after all opposing forces are considered. During the jump, the ground must exert a force greater than the jumper's weight to create a net upward force, resulting in acceleration. This concept is essential for calculating the exact force the ground exerts on the jumper at the moment of takeoff.
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Related Practice
Textbook Question
Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate A has mass mA, and crate B has mass mB. The coefficient of kinetic friction between each crate and the surface is μk. The crates are pulled to the right at constant velocity by a horizontal force F Draw one or more free-body diagrams to calculate the following in terms of mA, mB, and μk: (b) the tension in the rope connecting the blocks.

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Textbook Question

In another version of the 'Giant Swing' (see Exercise 5.50), the seat is connected to two cables, one of which is horizontal (Fig. E5.51). The seat swings in a horizontal circle at a rate of 28.0 rpm (rev/min). If the seat weighs 255 N and an 825-N person is sitting in it, find the tension in each cable.

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Textbook Question
A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)
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Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.62 m/s2. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?
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