The dimensions of `K` in the equation `W=1/2Kx^(2)` is

Options

A. `M^(1)L^(0)T^(-2)`
B. `M^(0)L^(1)T^(-1)`
C. `M^(1)L^(1)T^(-2)`
D. `M^(1)L^(0)T^(-1)`

Answer

To find the dimensions of \( K \) in the equation \( W = \frac{1}{2} K x^2 \), we can follow these steps:

### Step 1: Identify the quantities in the equation
In the equation, we have:
- \( W \) represents work (or energy)
- \( K \) is the spring constant (or stiffness)
- \( x \) represents distance (or displacement)

### Step 2: Write the dimensions of known quantities
1. **Work (W)**: The dimension of work is given by the formula \( W = F \cdot d \), where \( F \) is force and \( d \) is distance. The dimension of force \( F \) is \( [M L T^{-2}] \) (mass × acceleration). Therefore, the dimension of work is:
\[ [W] = [F] \cdot [d] = [M L T^{-2}] \cdot [L] = [M L^2 T^{-2}] \]

2. **Distance (x)**: The dimension of distance is simply:
\[ [x] = [L] \]

### Step 3: Substitute the dimensions into the equation
The equation can be rearranged to solve for \( K \):
\[ K = \frac{2W}{x^2} \]

### Step 4: Substitute the dimensions into the rearranged equation
Substituting the dimensions we found:
\[ [K] = \frac{2[M L^2 T^{-2}]}{[L]^2} \]

### Step 5: Simplify the dimensions
Now, we simplify the dimensions:
\[ [K] = \frac{[M L^2 T^{-2}]}{[L^2]} = [M L^{2-2} T^{-2}] = [M L^0 T^{-2}] = [M T^{-2}] \]

### Conclusion
Thus, the dimensions of \( K \) are:
\[ [K] = [M T^{-2}] \]