A small object of mass of 100 g moves in a circular path. At a given instant velocity of the object is `10hatim//s` and acceleration is `(20hati+10hatj)m//s^(2)`. At this instant of time, rate of change of kinetic energy (in `"kg m"^(2)s^(-3)`) of the object is

Answer

To find the rate of change of kinetic energy of the object, we can follow these steps:

### Step 1: Understand the Kinetic Energy Formula
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the object and \( v \) is its velocity.

### Step 2: Differentiate Kinetic Energy with Respect to Time
To find the rate of change of kinetic energy with respect to time, we differentiate the kinetic energy formula:
\[ \frac{d(KE)}{dt} = \frac{d}{dt}\left(\frac{1}{2} m v^2\right) \]
Using the chain rule, we get:
\[ \frac{d(KE)}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = m v \frac{dv}{dt} \]

### Step 3: Identify Given Values
From the problem:
- Mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (convert grams to kilograms)
- Velocity \( v = 10 \, \text{m/s} \)
- Acceleration \( a = (20 \hat{i} + 10 \hat{j}) \, \text{m/s}^2 \)

### Step 4: Calculate \( \frac{dv}{dt} \)
The acceleration \( a \) is the rate of change of velocity with respect to time:
\[ \frac{dv}{dt} = a = (20 \hat{i} + 10 \hat{j}) \, \text{m/s}^2 \]

### Step 5: Calculate the Dot Product
To find \( v \cdot \frac{dv}{dt} \), we need to compute the dot product:
- Velocity vector \( v = 10 \hat{i} \)
- Acceleration vector \( a = (20 \hat{i} + 10 \hat{j}) \)

Calculating the dot product:
\[ v \cdot a = (10 \hat{i}) \cdot (20 \hat{i} + 10 \hat{j}) = 10 \cdot 20 + 0 = 200 \]

### Step 6: Substitute Values into the Rate of Change of Kinetic Energy
Now substitute the values into the formula for the rate of change of kinetic energy:
\[ \frac{d(KE)}{dt} = m v \frac{dv}{dt} = 0.1 \, \text{kg} \cdot 10 \, \text{m/s} \cdot 200 \, \text{m/s}^2 \]
\[ \frac{d(KE)}{dt} = 0.1 \cdot 10 \cdot 200 = 200 \, \text{kg m}^2/\text{s}^3 \]

### Final Answer
The rate of change of kinetic energy of the object is:
\[ \frac{d(KE)}{dt} = 200 \, \text{kg m}^2/\text{s}^3 \]

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