When an object of mass m moves with velocity v along a line, its kinetic energy is given by K = -mv² . Assuming the mass m is constant but K and v are functions of time t that are not necessarily constant, find an equation relating the rate of change of kinetic energy to the rate of change of velocity . At what rate is the kinetic energy changing for an object with mass 20 kg moving with velocity -2 m/sec and acceleration -5 m/sec? dK

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**Kinetic Energy and Rate of Change**

**Problem Statement:**

When an object of mass \( m \) moves with velocity \( v \) along a line, its kinetic energy is given by the formula:

\[ K = \frac{1}{2} mv^2 \]

Assuming the mass \( m \) is constant, but both \( K \) and \( v \) are functions of time \( t \), find an equation relating the rate of change of kinetic energy \( \frac{dK}{dt} \) to the rate of change of velocity \( \frac{dv}{dt} \).

**Specific Question:**

At what rate is the kinetic energy changing for an object with:

- Mass = 20 kg
- Velocity = -2 m/sec
- Acceleration = -5 m/sec²?

**Solution Approach:**

1. **Differentiate the Kinetic Energy Formula:**

   Using the chain rule for differentiation:

   \[
   \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} mv^2 \right)
   \]
   
   Since \( m \) is constant, it becomes:

   \[
   \frac{dK}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = mv \frac{dv}{dt}
   \]

2. **Substitute Given Values:**

   For an object with:
   - \( m = 20 \, \text{kg} \)
   - \( v = -2 \, \text{m/sec} \)
   - \( \frac{dv}{dt} = -5 \, \text{m/sec}^2 \)

   The rate of change of kinetic energy is:

   \[
   \frac{dK}{dt} = 20 \times (-2) \times (-5) = 200 \, \text{Joules/sec}
   \]

Thus, the kinetic energy is changing at a rate of 200 Joules per second.
Transcribed Image Text:**Kinetic Energy and Rate of Change** **Problem Statement:** When an object of mass \( m \) moves with velocity \( v \) along a line, its kinetic energy is given by the formula: \[ K = \frac{1}{2} mv^2 \] Assuming the mass \( m \) is constant, but both \( K \) and \( v \) are functions of time \( t \), find an equation relating the rate of change of kinetic energy \( \frac{dK}{dt} \) to the rate of change of velocity \( \frac{dv}{dt} \). **Specific Question:** At what rate is the kinetic energy changing for an object with: - Mass = 20 kg - Velocity = -2 m/sec - Acceleration = -5 m/sec²? **Solution Approach:** 1. **Differentiate the Kinetic Energy Formula:** Using the chain rule for differentiation: \[ \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} mv^2 \right) \] Since \( m \) is constant, it becomes: \[ \frac{dK}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = mv \frac{dv}{dt} \] 2. **Substitute Given Values:** For an object with: - \( m = 20 \, \text{kg} \) - \( v = -2 \, \text{m/sec} \) - \( \frac{dv}{dt} = -5 \, \text{m/sec}^2 \) The rate of change of kinetic energy is: \[ \frac{dK}{dt} = 20 \times (-2) \times (-5) = 200 \, \text{Joules/sec} \] Thus, the kinetic energy is changing at a rate of 200 Joules per second.
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