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Check the correctness of the equation `W=1/2mv^(2)-1/2"mu"^(2)` using the dimensional analysis, where W is the work done, m is the mass of body,u-its initial velocity and v its final velocity.

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To check the correctness of the equation \( W = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \) using dimensional analysis, we will follow these steps: ### Step 1: Identify the dimensions of each term in the equation. 1. **Left-hand side (LHS)**: The term \( W \) represents work done. The dimensional formula for work is: \[ [W] = \text{Joule} = \text{Newton} \cdot \text{meter} = \text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \] Thus, the dimensions of work are: \[ [W] = [M][L^2][T^{-2}] \] ### Step 2: Analyze the right-hand side (RHS) of the equation. 2. **Right-hand side (RHS)**: The RHS consists of two terms: \( \frac{1}{2} mv^2 \) and \( \frac{1}{2} mu^2 \). - For the term \( \frac{1}{2} mv^2 \): - The mass \( m \) has dimensions \( [M] \). - The velocity \( v \) has dimensions \( [L][T^{-1}] \). - Therefore, \( v^2 \) has dimensions \( [L^2][T^{-2}] \). - Thus, the dimensional formula for \( mv^2 \) is: \[ [mv^2] = [M][L^2][T^{-2}] \] - Similarly, for the term \( \frac{1}{2} mu^2 \): - The initial velocity \( u \) also has dimensions \( [L][T^{-1}] \), and thus \( u^2 \) has dimensions \( [L^2][T^{-2}] \). - Therefore, the dimensional formula for \( mu^2 \) is: \[ [mu^2] = [M][L^2][T^{-2}] \] ### Step 3: Combine the terms on the RHS. 3. Now, we can combine the two terms on the RHS: \[ \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \] Since both terms have the same dimensions \( [M][L^2][T^{-2}] \), we can subtract them: \[ [\text{RHS}] = [M][L^2][T^{-2}] \] ### Step 4: Compare the dimensions of LHS and RHS. 4. Now we compare the dimensions of LHS and RHS: - LHS: \( [W] = [M][L^2][T^{-2}] \) - RHS: \( [\text{RHS}] = [M][L^2][T^{-2}] \) Since both sides have the same dimensions, we conclude that the equation is dimensionally correct. ### Final Conclusion: The equation \( W = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \) is dimensionally correct. ---
To check the correctness of the equation \( W = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \) using dimensional analysis, we will follow these steps: ### Step 1: Identify the dimensions of each term in the equation. 1. **Left-hand side (LHS)**: The term \( W \) represents work done. The dimensional formula for work is: \[ [W] = \text{Joule} = \text{Newton} \cdot \text{meter} = \text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2} \] ...
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