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Force F=[3x^2hati+4hatj] N where , x in ...

Force `F=[3x^2hati+4hatj]` N where , x in metres, acts on a particle. How much work is done on the particle as it moves from coordinates (2m,3m) to (3m, 0m)?

A

7J

B

5J

C

14 J

D

8 J

Text Solution

AI Generated Solution

The correct Answer is:
To find the work done on a particle by a force as it moves from one point to another, we can use the formula for work done, which is given by the dot product of the force vector and the displacement vector. Here’s a step-by-step solution to the problem: ### Step 1: Identify the Force Vector The force acting on the particle is given as: \[ \mathbf{F} = 3x^2 \hat{i} + 4 \hat{j} \] where \( x \) is in meters. ### Step 2: Determine the Initial and Final Positions The particle moves from the coordinates: - Initial position \( (x_1, y_1) = (2 \, \text{m}, 3 \, \text{m}) \) - Final position \( (x_2, y_2) = (3 \, \text{m}, 0 \, \text{m}) \) ### Step 3: Calculate the Displacement Vector The displacement vector \( \mathbf{dr} \) can be expressed as: \[ \mathbf{dr} = dx \hat{i} + dy \hat{j} \] where \( dx = x_2 - x_1 = 3 - 2 = 1 \, \text{m} \) and \( dy = y_2 - y_1 = 0 - 3 = -3 \, \text{m} \). ### Step 4: Substitute the Values into the Work Done Formula The work done \( W \) is given by the integral of the force along the path of displacement: \[ W = \int \mathbf{F} \cdot \mathbf{dr} \] ### Step 5: Express the Dot Product Substituting \( \mathbf{F} \) and \( \mathbf{dr} \): \[ W = \int (3x^2 \hat{i} + 4 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \] Calculating the dot product: \[ W = \int (3x^2 dx + 4 dy) \] ### Step 6: Set Up the Integrals We will integrate each term separately. The limits for \( x \) will be from 2 to 3 and for \( y \) from 3 to 0. 1. For the \( 3x^2 \) term: \[ W_1 = \int_{2}^{3} 3x^2 \, dx \] 2. For the \( 4 \) term: \[ W_2 = \int_{3}^{0} 4 \, dy \] ### Step 7: Calculate the Integrals 1. Calculate \( W_1 \): \[ W_1 = 3 \left[ \frac{x^3}{3} \right]_{2}^{3} = [27 - 8] = 19 \, \text{J} \] 2. Calculate \( W_2 \): \[ W_2 = 4 \left[ y \right]_{3}^{0} = 4(0 - 3) = -12 \, \text{J} \] ### Step 8: Combine the Results Now, combine the results of both integrals: \[ W = W_1 + W_2 = 19 - 12 = 7 \, \text{J} \] ### Final Answer The total work done on the particle as it moves from (2m, 3m) to (3m, 0m) is: \[ \boxed{7 \, \text{J}} \]
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