Problem 18 - College Physics

Question

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) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(\underline{W}\). what force does the ground exert on him or her during the jump?

Step-by-Step Solution

Answer

(a) Initial speed is approximately 3.43 m/s. (b) The diagram includes weight and normal force. (c) Force by ground equals weight plus any additional force needed for acceleration.

Step 1: Determine Variables and Formula

To solve part (a), we need to calculate the initial speed using kinematic equations. We know the maximum height (h) is 0.6 m, initial speed (v_0) is what we need to find, and final velocity at maximum height (v) is 0 m/s. The formula used is \[v^2 = v_0^2 + 2gh\]where g is the acceleration due to gravity (9.8 m/s²).

Step 2: Plug in Known Values

Substitute the known values into the equation:\[0 = v_0^2 + 2(-9.8\, \text{m/s}^2)(0.6\, \text{m})\]Simplify to find an expression for \(v_0\).

Step 3: Solve for Initial Velocity

From the equation, we solve for \(v_0\):\[0 = v_0^2 - 11.76\]\[v_0^2 = 11.76\]\[v_0 = \sqrt{11.76}\]Calculate \(v_0\) to find the initial speed, \(v_0 \approx 3.43 \text{ m/s}\).

Step 4: Create Free-Body Diagram

For part (b), draw a free-body diagram of the person during the jump. Show two forces: the weight of the person (\(\underline{W}\)) acting downward and the normal force \(F_n\), exerted by the ground, acting upward.

Step 5: Calculate Force by Ground

For part (c), use Newton's Second Law to determine the force exerted by the ground. The net force \(F_{\text{net}}\) during the jump phase is the difference between the force applied by the ground and the weight, so:\[F_n - \underline{W} = ma\]where \(a\) is the acceleration calculated from the initial speed over the jump distance (0.5 m). Rearrange to find \(F_n\).

Key Concepts

Initial velocity calculation

When jumping upwards, one important concept is calculating the initial velocity, which is the speed at which you leave the ground. To find this initial velocity, we often rely on kinematic equations, which are used to describe motion. Here, we know that the jumper reaches a height of 60 cm, which is 0.6 meters, and that the final velocity at this maximum height is 0 m/s, because the person momentarily stops before descending.
A key formula in kinematics is:

• \[v^2 = v_0^2 + 2gh\]

where \(v\) is the final velocity (0 m/s at the top), \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity (approximated as 9.8 m/s²), and \(h\) is the maximum height reached. In our case, we simplify the formula to solve for \(v_0\):

• \[v_0 = \sqrt{11.76}\]
• \(v_0 \approx 3.43 \mathrm{~m/s}\)

This means the person leaves the ground at an initial speed of approximately 3.43 meters per second.

Free-body diagram

A free-body diagram is a sketch used to show all the forces acting upon an object in a particular situation. In the case of the jumping person, the diagram serves as a visual representation of the mechanical interactions at play during the jump.
For this jump:

• The weight \((\underline{W})\) of the jumper acts downward due to gravity. This force is always present and depends on the mass of the person.
• The normal force \((F_n)\) acts upward. This is the force exerted by the ground to support the jumper, and it increases as the person speeds up during the jump to oppose the downward pull of gravity.

In a free-body diagram:

• The weight is an arrow pointing downwards.
• The normal force is a longer arrow pointing upwards because the ground must exert extra force to propel the person upwards.

Visualizing these forces helps in understanding the physical interactions and sets a foundation for calculating other important quantities like acceleration and net force.

Newton's Second Law

One of the fundamental principles of motion is Newton's Second Law of Motion, which provides a strong basis for understanding forces during movement. The law states that the net force acting on an object is equal to the product of its mass and acceleration:

• \[F_{\text{net}} = ma\]

In the jumping scenario, the net force during the jump is the result of the ground force minus the weight of the person:

• \[F_n - \underline{W} = ma\]

Here:

• \(F_n\) is the ground force, which is larger than the weight because it must not only counteract gravity but also propel the person upwards.
• \(a\) is the acceleration experienced during the jump, which can be determined from the change in speed over the distance the jumper's body travels.

Rearranging the equation allows us to solve for \(F_n\), showing us how much force the ground exerts on the jumper to make the leap possible.

Acceleration due to gravity

Acceleration due to gravity, often represented by the symbol \(g\), is a key concept when dealing with free fall or vertical jumps. This constant acceleration is approximately 9.8 meters per second squared \((\text{m/s}^2)\) on Earth's surface. Its role is crucial in calculating the motions of objects moving under the influence of this gravitational pull.
When a person jumps:

• Gravity is the force that pulls them back down, opposing their initial upward motion.
• During upward motion, gravity reduces the person's speed until they reach the apex of the jump where the speed becomes zero.
• It then accelerates the person back towards the ground.

Understanding \(g\) helps us solve for other variables like initial velocity and is integral to using kinematic equations accurately.