Problem 14 - College Physics

Question

A \(65.0-\mathrm{kg}\) basketball player jumps vertically and leases the floor with a velocity of \(1.80 \mathrm{~m} / \mathrm{s}\) upward. (a) What impulse does the player experience? (b) What force does the floor exert on the player before the jump? (c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for \(0.450\) s during the jump?

Step-by-Step Solution

Answer

The impulse experienced by the player is \(117.0 \mathrm{~kg} ~\mathrm{m} / \mathrm{s}\) upward. The force exerted by the floor on the player before the jump is \(637.0 \mathrm{~N}\) upward, and the total average force exerted by the floor on the player during the jump is \(1060.0 \mathrm{~N}\) upward.

Step 1: Calculate the Impulse Experienced by the Player

Impulse is the change in momentum of an object when a force is applied. It is given by the product of the mass of the object and the change in velocity. In this case, the player's initial momentum was zero (as he was at rest), and the final momentum is the product of his mass and his final velocity. Therefore, the impulse \(J\) experienced by the player can be calculated as: \(J=m \cdot \Delta v= m \cdot v\), where \(m=65.0 \mathrm{~kg}\) (mass of the player) and \(v=1.80 \mathrm{~m} / \mathrm{s}\) (velocity of the player).

Step 2: Calculate the Force Exerted by the Floor on the Player Before the Jump

This force is simply the weight of the player, which is the product of the mass of the player and gravity. \(F=m \cdot g\), where \(m=65.0 \mathrm{~kg}\) (mass) and \(g \approx 9.8 \mathrm{~m} / \mathrm{s}^2\) (acceleration due to gravity).

Step 3: Compute the Total Average Force Exerted by the Floor on the Player During the Jump

The total average force exerted on the player by the floor during the jump is the overall impulse divided by the time the player was in contact with the floor. \(F_{avg} = \frac{J}{t}\), where \(J\) is the impulse calculated in Step 1 and \(t=0.450 \mathrm{~s}\) (contact time with the floor).

Key Concepts

Momentum Change in Physics

Momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It is the product of an object's mass and its velocity. To understand momentum change, imagine a basketball player making a jump. Initially, while standing still, the player's momentum is zero. Upon jumping, the player acquires momentum equal to their mass multiplied by the upward velocity.

Momentum change, also known as impulse, occurs when an external force acts on an object, leading to a change in its velocity. In the case of the basketball player who weighs 65.0 kg and leaves the ground with a velocity of 1.80 m/s, the momentum change can be found by subtracting the initial momentum from the final momentum. Since the initial momentum was zero, the momentum change equals the final momentum, which is the product of the mass and the final velocity, \( \Delta p = m \cdot \Delta v \), where \( \Delta v \) is the change in velocity. This gives us a clear insight into how the player's motion is influenced by the act of jumping.

Force Calculation in Physics

In the context of a basketball player jumping, force calculation involves determining the force exerted by the floor on the player before the jump. This force must at least equal the gravitational force acting on the player, also known as the player's weight. It is calculated using the equation \( F = m \cdot g \), where \( m \) is the mass of the player and \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)).

Using this simple formula, we find that the weight of the player, or the force exerted by the floor when the player is static, is the product of their mass (65.0 kg) and the acceleration due to gravity. This force represents a baseline that needs to be overcome by additional force to allow for the player's upward movement during the jump. Recognizing this force is essential for understanding the forces at play in many physical scenarios, from sports to everyday actions.

Average Force and Its Significance

Average force is a concept in physics used to describe the constant force that would produce the same effect on an object's motion as the actual force exerted over a period of time. When considering the basketball player's jump, the average force exerted by the floor can be calculated by dividing the impulse, or the momentum change, by the time interval during which the player is in contact with the floor. The formula \( F_{avg} = \frac{J}{t} \) allows us to calculate this average force, where \( J \) is the change in momentum, and \( t \) is the contact time.

This average force, unlike the instantaneous force felt at any given moment, provides a practical and comprehensive understanding of how the force is distributed over the duration of the contact. The average force calculation is valuable in designing sports equipment, understanding athletic performance, and even in engineering applications where distributed forces need to be considered.