A small object of mass of 100 g moves in a circular path . At a given instant velocity of the object is `10hati m s^(-1)` and acceleration is `(20hati+10hatj) m s^(-2)`. At this instant of time. the rate of change of kinetic energy of the object is

Options

A. `200 "kg m"^2 s^(-2)`
B. `300 "kg m"^2 s^(-2)`
C. `10000 "kg m"^2 s^(-2)`
D. `20 "kg m"^2 s^(-2)`

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, we differentiate \( KE \) with respect to time \( t \):
\[ \frac{d(KE)}{dt} = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) = \frac{1}{2} m \frac{d(v^2)}{dt} \]
Using the chain rule, we can express \( \frac{d(v^2)}{dt} \) as:
\[ \frac{d(v^2)}{dt} = 2v \cdot \frac{dv}{dt} \]
Thus, we have:
\[ \frac{d(KE)}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = m v \cdot \frac{dv}{dt} \]

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

We need to find \( \frac{dv}{dt} \), which is equal to the acceleration \( a \).

### Step 4: Calculate the Rate of Change of Kinetic Energy
Substituting the values into the formula:
\[ \frac{d(KE)}{dt} = m v \cdot a \]
Calculating \( v \cdot a \):
\[ v \cdot a = (10 \hat{i}) \cdot (20 \hat{i} + 10 \hat{j}) = 10 \cdot 20 + 0 = 200 \, \text{m}^2/\text{s}^2 \]
Now substituting \( m = 0.1 \, \text{kg} \):
\[ \frac{d(KE)}{dt} = 0.1 \cdot 200 = 20 \, \text{W} \]

### Final Answer
The rate of change of kinetic energy of the object is:
\[ \frac{d(KE)}{dt} = 20 \, \text{W} \]