Problem 21 - College Physics

Question

\(\bullet$$\bullet\) 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 reason- able 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) Make 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

Initial speed: 3.43 m/s. Ground force: 2.2 times the weight.

Step 1: Understand the Problem

We need to calculate the initial speed with which a person leaves the ground to reach a maximum height of 60 cm during a jump. Then, we'll draw a free-body diagram of the forces acting on the person during the jump, and calculate the force the ground exerts in terms of the person's weight, W.

Step 2: Calculate Initial Speed (Part a)

We use the kinematic equation for motion under gravity to find the initial velocity \( v_i \):\[ v_f^2 = v_i^2 + 2a \, s \]where \( v_f = 0 \) (at the maximum height), \( a = -9.8 \text{ m/s}^2 \) (acceleration due to gravity), and \( s = 0.6 \text{ m (60 cm)} \). Rearranging gives:\[ 0 = v_i^2 - 2 \times 9.8 \times 0.6 \]\[ v_i^2 = 11.76 \]\[ v_i = \sqrt{11.76} \approx 3.43 \text{ m/s} \] Thus, the initial speed is approximately 3.43 m/s.

Step 3: Free-Body Diagram (Part b)

During the jump, the main forces acting on the person are the gravitational force \( F_{gravity} = mg \) downward and the normal force \( F_{normal} \) upward exerted by the ground. The diagram would show the person with these two forces marked in opposite directions. The ground force is greater than the gravitational force to provide the upward acceleration.

Step 4: Calculate Ground Force (Part c)

Assuming the person's body rises 0.5 m during the jump and using the calculated initial velocity, we find the acceleration \( a \) using the equation for initial velocity with acceleration:\[ v_i = at \]or:\[ v_f^2 = v_i^2 + 2a \, s\]Here, \( v_f = v_i \, (from rest position to initial speed), v_i = 3.43 \text{ m/s}, s = 0.5 \text{ m} \), hence:\[ 3.43^2 = 0 + 2a \times 0.5 \]\[ 11.76 = a \]When solving for the net force, \( F_{net} = ma = F_{normal} - mg \), we find:\[ F_{normal} = ma + mg = m(11.76 + 9.8) = 21.56m \]Expressed in terms of weight \( W \), where \( W = mg \), the force is:\[ F_{normal} = 2.2W \]

Step 5: Summarize Conclusions

The initial speed is approximately 3.43 m/s. The forces acting on the person during the jump include gravity and the normal force, which is 2.2 times the person's weight due to additional acceleration provided by the ground.

Key Concepts

Kinematic equations

Kinematic equations are a set of formulas that describe the motion of objects. They are particularly useful when you are dealing with uniform acceleration, like gravity. The equation we used in this problem is one of the most common kinematic equations:

• \( v_f^2 = v_i^2 + 2a \, s \)

This formula helps us determine the initial speed \( v_i \) when an object reaches a certain height. Here, the final velocity \( v_f \) is zero because the person momentarily stops at the maximum height before descending.
The acceleration \( a \) is negative due to gravity, which always acts downward. By rearranging the equation to solve for \( v_i \), we were able to calculate how fast the jumper must launch off the ground to reach a height of 60 cm.
Mastering kinematic equations is crucial because it allows you to break down motions into understandable parts, making problem-solving more manageable.

Free-body diagrams

Free-body diagrams are visual tools used in physics to illustrate the forces acting on a single object. By focusing purely on the object of interest and mapping out all the forces, these diagrams simplify complex problems.
For the jumping person, the main forces to consider are:

• Gravitational force (\( F_{gravity} \)), which pulls the person downward.
• Normal force (\( F_{normal} \)), which the ground exerts upward against the jumper.

During a jump, the normal force exceeds the gravitational force, providing the upward acceleration needed to leave the ground. The free-body diagram effectively highlights this excess force by showing the imbalance required for upward motion.
By focusing on these force interactions, you gain valuable insight into how objects interact with their surroundings and how motion results from these interactions.

Forces during motion

Understanding forces during motion is essential for analyzing and predicting how objects behave. In a jumping scenario, two main forces play significant roles: the gravitational force and the normal force exerted by the ground.
The gravitational force acts to pull the person back towards the earth, which is constant and equals \( mg \) where \( m \) is mass and \( g \) is the acceleration due to gravity.
The normal force, on the other hand, is a reaction force from the ground. During a jump, the ground must exert a force greater than gravity to propel the person upwards. This makes the normal force around 2.2 times the weight of the jumper, which is necessary to accelerate and achieve the desired height.

• The net force \( F_{net} \) leads to the acceleration \( a \) necessary for the jump.
• It emphasises Newton's second law, \( F = ma \), where the net force creates acceleration.

Understanding these forces in motion showcases how they work together to result in the motion experienced in activities like jumping.