Problem 21 - University Physics with Modern Physics

Question

Force During a Jump. An average person can reach a maximum height of about 60 \(\mathrm{cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 \(\mathrm{cm}\) .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 \(\mathrm{cm} ?\) (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?

Step-by-Step Solution

Answer

(a) 3.43 m/s (b) Forces: Normal upward, weight downward (c) 2.20 times weight

Step 1: Determine the initial velocity

We will use the kinematic equation \(v^2 = u^2 + 2as\) to find the initial speed \(u\) needed to reach a height of 60 cm, where the final velocity \(v = 0\) at the peak of the jump, and \(a = -9.8\, \text{m/s}^2\) (acceleration due to gravity). The height \(s\) is 0.6 m.Rearranging the equation gives: \(u^2 = v^2 - 2as\). Substituting the known values: \(u^2 = 0 - 2(-9.8)(0.6)\). Calculating this gives \(u^2 = 11.76\), thus \(u = \sqrt{11.76} = 3.43\, \text{m/s}\).

Step 2: Create a free-body diagram

The free-body diagram should show the forces acting on the person during the jump. These forces include the gravitational force (the person's weight \(w\) acting downward) and the normal force from the ground acting upward. The normal force is greater than the weight during the jump to provide the net upward force needed to accelerate the person.

Step 3: Calculate the net force during the jump

The net force can be calculated using Newton's second law, \(F_{net} = ma\). But first, we need to determine the acceleration \(a\) using the equation \(v^2 = u^2 + 2as\), with the upward distance during the jump as 0.5 meters and initial velocity \(u=0\) at the crouch.Rearranging for acceleration: \(a = \frac{v^2 - u^2}{2s}\). Substitute \(v = 3.43\, \text{m/s}\), \(u = 0\), \(s = 0.5\): \(a = \frac{(3.43)^2 - 0}{2(0.5)} = 11.76\, \text{m/s}^2\).

Step 4: Calculate the ground force

With the acceleration found in Step 3, apply Newton's second law, \(F_{ground} = ma + w\) (where \(w\) is the person's weight, \(w = mg\)). Knowing \(a = 11.76\, \text{m/s}^2\) and the acceleration due to gravity \(g = 9.8\, \text{m/s}^2\), the force from the ground is: \[ F_{ground} = m(11.76) + mg = m(11.76 + 9.8) = m(21.56) \]Thus the ground force is \(2.20w\), where \(w = mg\).

Key Concepts

free-body diagram

A free-body diagram is a simple illustration used to visualize the forces acting upon an object. In this exercise, the free-body diagram for a person jumping is an essential part to understand how forces interact during movement.

When considering this jump, two primary forces appear on the diagram:

• The gravitational force, which is the person's weight (\(w\)), acting downwards.
• The ground's normal force, pointing upwards, which exceeds the gravitational force to propel the person upwards.

These forces show that the net force results in an upward acceleration, allowing the person to leave the ground. Remember, for every upward motion, there should be a force greater than the downward force acting against gravity. This balance is crucial for understanding the dynamics of jumping or any vertical motion.

Newton's second law

Newton's second law of motion is fundamental to understanding the forces in a jump. It is represented by the equation \(F_{net} = ma\), where \(F_{net}\) is the net force applied, \(m\) is mass, and \(a\) is the acceleration.

When a person jumps, the muscles provide extra force to push off the ground. This additional force results in acceleration beyond what gravity alone would cause. By applying Newton's second law, one can determine the net force necessary to achieve the upward motion. In this exercise, we see that the acceleration achieved during the jump is found to be \(11.76 \,\text{m/s}^2\). This result is crucial for calculating the ground reaction force, which must be greater than the gravitational pull to lift the person upwards. This equation beautifully ties the concepts of force, mass, and acceleration together in motion examples like jumping.

initial velocity

Initial velocity is the speed at which an object starts its motion. In the context of this exercise, it refers to the speed the jumper needs to leave the ground to achieve a certain height.

To calculate the initial velocity required to reach 60 cm, we use the kinematic equation \(v^2 = u^2 + 2as\). Here, the final velocity \(v\) is zero at the highest point of the jump, \(a\) is the acceleration due to gravity (\(-9.8\,\text{m/s}^2\)), and \(s\) is the displacement, 60 cm or 0.6 meters.
After rearranging the formula, you find that \(u\), the initial velocity, is \(3.43 \,\text{m/s}\). This calculation is crucial because it tells how fast the person must initially leave the ground to reach the desired height and battle gravitational forces effectively.

acceleration due to gravity

Acceleration due to gravity is a constant that significantly influences motion on Earth. It is usually symbolized by \(g\) and is approximately \(9.8\,\text{m/s}^2\) near the Earth's surface.

This constant defines how fast an object will accelerate downwards when dropped or how strongly it pulls objects back to Earth. In this exercise, gravity serves as the force that the jumper must overcome to reach the desired height. During the jump, while the person pushes off, the force applied must counteract the gravitational pull to achieve the initial upward motion. Knowing the value of \(g\) is fundamental when using kinematic equations, as it helps us calculate how much speed or force we need to perform tasks like jumping or throwing an object upwards.