Question

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work\(-\)energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

Short answer

(a) The initial speed is approximately 30.3 m/s, and (b) the maximum height is 39.5 m.

Step-by-step solution

Understand the Problem

We need to find the initial speed of the rock (when it's just leaving the ground) and the maximum height it reaches using the work-energy theorem.

Identify Known Values

The rock's weight is 3.00 N, its velocity at 15.0 m height is 25.0 m/s, and we need to find its speed at the ground and its maximum height.

Calculate the Mass of the Rock

Use the gravitational force formula: \[ F = m imes g \] where \( F = 3.00 \, \text{N} \), and \( g = 9.81 \, \text{m/s}^2 \). Solve for \( m \): \[ m = \frac{3.00}{9.81} \, \text{kg} \approx 0.306 \text{ kg} \].

Use Work-Energy Theorem for Initial Speed

The work-energy theorem states that the change in kinetic energy is equal to the change in potential energy. Starting from ground level to 15 meters: \[ \frac{1}{2} m v^2_1 + mgh = \frac{1}{2} m v^2_2 \]where \( h = 15 \) m, \( v_2 = 25 \) m/s, and gravity \( g = 9.81 \) m/s². Solve for \( v_1 \).

Solve for Initial Speed \(v_1\) Equation

Input known values into the equation: \[ \frac{1}{2} m v_1^2 + m \times 9.81 \times 15 = \frac{1}{2} \times 0.306 \times 25^2 \]After calculations, solve for \( v_1 \) to find the initial speed.

Calculate Rock's Maximum Height

To find the maximum height, use kinetic energy at maximum height equal to zero and potential energy. \[ \frac{1}{2} m v_1^2 = mgh_{max} \]Solve for \( h_{max} \) after obtaining \( v_1 \).

Solve for Maximum Height Equation

After finding \( v_1 \), substitute back into \[ h_{max} = \frac{v_1^2}{2g} \].

Key concepts

Kinetic Energy

Kinetic energy is the energy of motion. When you throw an object, like a rock, into the air, it has kinetic energy because it is moving. The formula for kinetic energy is\[ KE = \frac{1}{2}mv^2 \]where:

• \( m \) is the mass of the object, and
• \( v \) is its velocity.

In the exercise, as the rock is thrown upwards, it initially has a certain kinetic energy based on its speed and mass. Calculating this helps us find out the speed of the rock when it first leaves the ground. This is important because knowing the initial kinetic energy allows us to understand how much energy the rock had due to motion right from its launch point.

Potential Energy

Potential energy is stored energy due to position. For an object that is lifted against gravity, potential energy comes from height. The formula for potential energy is \[ PE = mgh \]where:

• \( m \) is the mass,
• \( g \) is the acceleration due to gravity \( (9.81 \, \text{m/s}^2) \), and
• \( h \) is the height above the ground.

When the rock reaches a height of 15 m, it gains potential energy. The higher it goes, the more potential energy it stores. This shift from kinetic energy to potential energy explains how energy is conserved as the rock rises. Understanding this relationship helps in predicting the rock's future behavior, such as reaching its maximum height.

Gravitational Force

Gravitational force is a fundamental force that attracts two bodies towards each other. For objects near the Earth's surface, this force gives the object weight. The gravitational force acting on the rock can be calculated using \[ F = mg \]where:

• \( F \) is the force or weight (in newtons),
• \( m \) is the mass, and
• \( g \) is the gravitational acceleration.

In this exercise, knowing the weight of the rock (3.00 N) allows us to determine its mass, important for all subsequent energy calculations. Gravitational force not only gives the rock weight but also affects how high it can go when thrown into the air. This force constantly acts as the rock moves upwards and eventually pulls it downwards, influencing both kinetic and potential energy.

Maximum Height

The maximum height of the rock is the highest point it reaches before it starts descending. At this point, all the initial kinetic energy has been converted into potential energy.
To find the maximum height, we can use the energy conversion principles illustrated in \[ \frac{1}{2} m v_1^2 = mgh_{max} \]. This means the maximum height \( h_{max} \) can be calculated by rearranging the equation to \[ h_{max} = \frac{v_1^2}{2g} \].
At maximum height, the velocity of the rock is 0 m/s because it momentarily stops before gravity pulls it back down. This concept helps us understand how energy is transferred from motion to potential due to the height increase. It's also a key component of the work-energy theorem, providing insight into how energy dictates height and motion.