Problem 9 - Applied Physics

Question

A box is dropped \(3.60 \mathrm{~m}\) to the ground. What is its speed as it hits the ground?

Step-by-Step Solution

Answer

The speed of the box as it hits the ground is approximately 8.40 m/s.

Step 1: Identify the known values

The height from which the box is dropped is \( h = 3.60 \, \mathrm{m} \). The initial speed \( v_0 = 0 \, \mathrm{m/s} \) because the box is dropped (not thrown). The acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \).

Step 2: Choose the appropriate physics equation

To find the final velocity \( v \) of the box, use the kinematic equation: \[ v^2 = v_0^2 + 2gh \]This equation relates the initial and final velocities, acceleration due to gravity, and distance fallen.

Step 3: Substitute the known values into the equation

Since \( v_0 = 0 \, \mathrm{m/s} \), the equation simplifies to: \[ v^2 = 0 + 2 \times 9.81 \, \mathrm{m/s^2} \times 3.60 \, \mathrm{m} \]Calculating further, this becomes:\[ v^2 = 70.632 \, \mathrm{m^2/s^2} \]

Step 4: Solve for the final velocity

Take the square root of both sides to solve for \( v \):\[ v = \sqrt{70.632} \, \mathrm{m/s} \]\[ v \approx 8.40 \, \mathrm{m/s} \]

Step 5: Conclude with the result

Thus, the speed of the box as it hits the ground is approximately \( 8.40 \, \mathrm{m/s} \).

Key Concepts

Projectile Motion

When an object is in motion under the influence of gravity alone, and moves along a curved path, it is often described as being in "projectile motion." In the given exercise, though the box is initially at rest and falls directly downwards, this scenario is a simplified case related to projectile motion. Please note that our case is vertical motion. In general, projectile motion has two components:

• Horizontal motion - uniform velocity
• Vertical motion - accelerated motion due to gravity

However, when an object is dropped and not thrown, like the box here, only vertical motion is considered. This situation simplifies the analysis since the object moves along a straight line and is only subjected to gravity.
Understanding projectile motion is crucial as it helps predict how an object behaves when launched into the air freely, which encompasses a wide range of applications from sports to engineering.

Gravitational Acceleration

Gravitational acceleration, denoted by the symbol \( g \), is the acceleration of an object caused by Earth's gravity. Its standard value is \( 9.81 \, \mathrm{m/s^2} \), meaning any object, in free fall, will increase its speed by \( 9.81 \, \mathrm{m/s} \) every second, neglecting air resistance.

In our exercise, the box is dropped from rest, and only gravity influences its motion. Therefore, gravitational acceleration is key to understanding how the speed of the box changes over time and distance. It's worthwhile noting that the value of \( g \) might differ slightly depending on geographical location due to variance in Earth's surface. However, \( 9.81 \, \mathrm{m/s^2} \) is commonly used for standard calculations.

Gravitational acceleration is a fundamental concept in physics and underpinning various phenomena, from the simple act of dropping an object to more complex systems like planetary orbits.

Velocity Calculation

Calculating the velocity of an object during motion is a fundamental aspect of kinematics, especially when understanding how velocity changes over time due to external factors like gravity. In the given exercise, we use a specific kinematic equation to determine the final velocity of a dropped box:

\[ v^2 = v_0^2 + 2gh \]
This equation connects the initial velocity \( v_0 \), gravitational acceleration \( g \), and the height \( h \), to find the final velocity \( v \) before the box hits the ground.

• First, acknowledge the initial velocity \( v_0 \): Since the box is dropped, \( v_0 = 0 \).
• Substitute the known values: Simply plug the height and gravitational acceleration into the equation.
• Calculate \( v \): Finally, solve for \( v \) by taking the square root of the result.

In step 4, you see how easily numeric substitution turns into real data. The calculated result of approximately \( 8.40 \, \mathrm{m/s} \) allows us to understand how fast the box is moving right before it touches the ground.