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Chapter 5: Problem 21

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. (a) With what initial speed does the person leave the ground to reach a height of 60 cm? (b) Draw a free-body diagram of the person 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?

Short Answer

Expert verified
(a) Initial speed is 3.43 m/s. (b) Free-body diagram: upward normal force, downward gravitational force. (c) Force exerted by ground is 21.56 times the weight.

Step by step solution

01

Understand the Problem

We need to calculate the initial speed of a person jumping upwards to reach a peak height of 60 cm. This will involve using physics concepts such as kinematics. Additionally, we need to create a free-body diagram and calculate the force exerted by the ground.
02

Use the Kinematic Equation for Height

The height reached (\( h = 60 \text{ cm} = 0.60 \text{ m} \)) is related to the initial velocity (\( v_0 \)) and acceleration due to gravity (\( g = 9.8 \text{ m/s}^2 \)) with the equation: \[ h = \frac{v_0^2}{2g}\]. Rearrange to solve for \( v_0 \): \[v_0 = \sqrt{2gh}\].
03

Calculate the Initial Speed

Insert the values into the rearranged equation: \[v_0 = \sqrt{2 \times 9.8 \times 0.60} = \sqrt{11.76} \approx 3.43 \text{ m/s}\]. The initial speed the jumper needs to leave the ground is approximately 3.43 m/s.
04

Draw a Free-Body Diagram

In a free-body diagram for the person during the jump, consider forces acting on the person: the gravitational force (\( mg \)) acting downwards, and the normal force from the ground acting upwards. Since the person is accelerating upwards during the jump, the normal force is larger than gravitational force.
05

Calculate Force Exerted by the Ground

Using Newton's second law (\( F = ma \)), and knowing that acceleration (\( a \) is upward), first find acceleration using \( v_f = v_0 + at \). As the person leaves the ground at 3.43 m/s and initially moves 0.5 m: \[ v_f^2 = v_0^2 + 2ad \rightarrow 0 = 3.43^2 + 2a(0.5) \rightarrow a = -3.431^2 / (2 \times 0.5) = 11.76 \text{ m/s}^2 \]. Thus, the total force is: \[ F_g = ma + mg = m(11.76 + 9.8) = 21.56m \], with the ground exerting a force of \( 21.56 \times \text{weight}(w) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Velocity Calculation
When a person jumps straight up, we need to calculate the speed needed for them to leave the ground and reach a particular height. This speed is known as the initial speed or initial velocity. To do this, we use a fundamental equation from kinematics. This equation tells us how the height a person can jump is related to how fast they leave the ground and how gravity pulls them back down.

The key formula for this is:
  • \[ h = \frac{v_0^2}{2g} \]
Here, \( h \) is the height reached (60 cm or 0.6 m for this exercise), \( v_0 \) is the initial velocity, and \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏). Rearranging the formula to solve for \( v_0 \) gives us:
  • \[ v_0 = \sqrt{2gh} \]
By plugging in our values, we find:
  • \( v_0 = \sqrt{2 \times 9.8 \times 0.6} = \sqrt{11.76} \approx 3.43 \text{ m/s} \)
This means the person's initial speed must be approximately 3.43 meters per second to reach 60 cm off the ground.
Free-Body Diagram
A free-body diagram is a simple drawing that shows all the forces acting on an object. For a person jumping straight up, this helps us see and understand the forces involved. In physics, the key forces are the gravitational force that pulls the person down and the normal force exerted by the ground pushing them up.

Here鈥檚 what a free-body diagram would typically include:
  • **Gravitational Force (Weight):** The person has a weight because of their mass and gravity, represented as \( mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity.
  • **Normal Force:**This force is exerted upwards by the ground. Since the person is accelerating upwards during the jump, this normal force is greater than the gravitational force.
Drawing these forces correctly is crucial because it illustrates the difference between when the person is standing still and when they're launching into the air. The extra upward force explains the upward motion and why the person can leave the ground.
Force Exerted by the Ground
When the person jumps, the ground must exert an additional force to overcome gravity and propel them upwards. This force can be calculated using Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object times its acceleration (\( F = ma \)).

To find the force exerted by the ground:
  • First, calculate the acceleration required to reach the initial speed when jumping. The equation \( v_f = v_0 + at \) will help here, taking into account distance covered.
  • By inserting known values, you find:\[ v_f^2 = v_0^2 + 2ad \rightarrow 0 = 3.43^2 + 2a(0.5) \] which results in \( a = 11.76 \text{ m/s}^2 \).
  • Now, consider the total force exerted on the person during the jump:\[ F_g = ma + mg = m(11.76 + 9.8) = 21.56m \]
The force exerted by the ground must be substantial enough to not only support the person's weight but also to provide the acceleration necessary to lift them skyward, essentially doing much more than just countering gravity.

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Most popular questions from this chapter

You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled 192 ft before stopping, and the coefficient of kinetic friction between the road and his tires was 0.750. He was charged with speeding in a 45-mi/h zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t) = At + Bt^2\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s\(^2\) at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m>s. (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

Some sliding rocks approach the base of a hill with a speed of 12 m/s. The hill rises at 36\(^\circ\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and 0.65, respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

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.)

A box is sliding with a constant speed of 4.00 m/s in the \(+x\)-direction on a horizontal, frictionless surface. At \(x =\) 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between \(x =\) 0 and \(x =\) 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400. (a) What is the \(x\)-coordinate of the point where the box comes to rest? (b) How much time does it take the box to come to rest after it first encounters the rough patch at \(x =\) 0?
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