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 solve the problem of finding the rate of change of kinetic energy of the object, we can follow these steps:

### Step 1: Understand the Given Data
We have:
- Mass of the object, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Velocity of the object, \( \vec{v} = 10 \hat{i} \, \text{m/s} \)
- Acceleration of the object, \( \vec{a} = 20 \hat{i} + 10 \hat{j} \, \text{m/s}^2 \)

### Step 2: Recall the Formula for Rate of Change of Kinetic Energy
The rate of change of kinetic energy can be expressed as the power, which is given by the formula:
\[ \text{Power} = \vec{F} \cdot \vec{v} \]
Where \( \vec{F} \) is the net force acting on the object. According to Newton's second law, \( \vec{F} = m \vec{a} \).

### Step 3: Calculate the Force
Using the mass and acceleration:
\[ \vec{F} = m \vec{a} = 0.1 \, \text{kg} \cdot (20 \hat{i} + 10 \hat{j}) \, \text{m/s}^2 = (2 \hat{i} + 1 \hat{j}) \, \text{N} \]

### Step 4: Calculate the Dot Product of Force and Velocity
Now, we need to find the dot product \( \vec{F} \cdot \vec{v} \):
\[ \vec{F} \cdot \vec{v} = (2 \hat{i} + 1 \hat{j}) \cdot (10 \hat{i}) = 2 \cdot 10 + 1 \cdot 0 = 20 \, \text{W} \]

### Step 5: Conclusion
The rate of change of kinetic energy of the object is:
\[ \frac{dKE}{dt} = 20 \, \text{W} \]

### Final Answer
The rate of change of kinetic energy of the object is \( 20 \, \text{W} \).

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