Question

Kinetic Energy The kinetic energy \(K\) of a moving object varies directly with its mass \(m\) and the square of its velocity \(v\). If an object weighing 25 kilograms and moving with a velocity of 10 meters per second has a kinetic energy of 1250 joules, find its kinetic energy when the velocity is 15 meters per second.

Short answer

The kinetic energy is 2812.5 joules.

Step-by-step solution

- Write the formula for kinetic energy

The kinetic energy (\text{K}) of an object is given by the formula: \[ K = \frac{1}{2}mv^2 \] where \( K \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity.

- Plug in the given values to find the constant of variation

Given that for an object with mass \( m = 25 \text{kg} \) and velocity \( v = 10 \text{m/s} \), the kinetic energy \( K = 1250 \text{J} \), plug in these values into the kinetic energy formula: \[ 1250 = \frac{1}{2} \times 25 \times 10^2 \].

- Solve for the constant

Simplify the equation to find the constant: \[ 1250 = \frac{1}{2} \times 25 \times 100 \] \[ 1250 = \frac{1}{2} \times 2500 \] \[ 1250 = 1250 \] This confirms the constant. Since it checks out, we can proceed to the next part using the same constant.

- Calculate the new kinetic energy with velocity 15 m/s

Now, find the new kinetic energy \( K \) when the velocity \( v = 15 \text{m/s} \). Use the formula again: \[ K = \frac{1}{2} \times 25 \times 15^2 \].

- Solve the equation

Calculate the value: \[ K = \frac{1}{2} \times 25 \times 225 \] \[ K = 12.5 \times 225 \] \[ K = 2812.5 \text{J} \].

Key concepts

Direct Variation

In math, direct variation describes a relationship between two variables in which one is a constant multiple of the other. This means when one variable changes, the other changes in proportion to the first. In our case, the kinetic energy (\text{K}) varies directly with both the mass (\text{m}) of an object and the square of its velocity (\text{v}). This relationship is expressed through the formula:

• \[ K = \frac{1}{2}mv^2 \]

This equation tells us that if the velocity or mass of the object increases, the kinetic energy will also increase proportionately. By understanding direct variation, we can predict how changes in one quantity will impact another, making it a powerful tool in solving physcis problems involving kinetic energy.

Mass and Velocity

Mass and velocity are crucial when discussing kinetic energy. Mass refers to the amount of matter in an object and is measured in kilograms (kg). Velocity, on the other hand, is the speed of the object in a given direction, measured in meters per second (m/s). When considering kinetic energy, both mass and the square of the velocity play a significant role. According to the formula, kinetic energy will increase if you increase either the mass or the velocity. More specifically:

• Increasing the mass (\text{m}) causes a direct increase in kinetic energy by the same factor.
• Increasing the velocity (\text{v}) causes the kinetic energy to increase by the square of the velocity increase.

For example, doubling the velocity results in a fourfold increase in kinetic energy, since \[ (2v)^2 = 4v^2 \]. Understanding these relationships helps in solving problems where you need to find the kinetic energy given changes in mass or velocity.

Solving Equations

Solving equations is essential when working with formulas. In our exercise, we follow these steps:

• First, understand the given values: mass (\text{m}) = 25 kg, initial velocity (\text{v}) = 10 m/s, and initial kinetic energy (\text{K}) = 1250 joules.
• Plug these values into the kinetic energy formula: \[ 1250 = \frac{1}{2} \times 25 \times 10^2.\]
• Simplify to find the specific constant, confirming \[ 1250 = \frac{1}{2} \times 2500.\]
• Using the same formula, find the new kinetic energy for the updated velocity (\text{v}) = 15 m/s: \[ K = \frac{1}{2} \times 25 \times 15^2.\]
• Calculate to find the new kinetic energy value: \[ K = 2812.5 \text{J} \].

Always ensure each step is clearly understood and verified, as solving equations involves understanding the relationship between variables and constants. This methodical approach will help consistently arrive at accurate solutions.