Work Done by a Variable Force
Question: A particle experiences a variable force \[ \vec{F}(x,y) = 3x^2\,\hat{i} + 2y\,\hat{j} \quad \text{(in Newtons)} \] in a horizontal \(x\text{–}y\) plane. If the particle moves from point \((2,1)\) to point \((3,2)\), then the change in its kinetic energy (equal to the work done) is:
- (1) 22.0 J
- (2) 12.5 J
- (3) 25.0 J
- (4) 0 J
Select the Correct Option
Detailed Step‑by‑Step Explanation
Step 1: The change in kinetic energy equals the work done on the particle when it moves from the initial point to the final point: \[ \Delta KE = W = \int_C \vec{F} \cdot d\vec{r}. \] Since the force is conservative (its curl is zero), the work is path independent.
Step 2: A convenient path is the straight line from \((2,1)\) to \((3,2)\). We parameterize the path by letting: \[ x = 2 + t,\quad y = 1 + t,\quad \text{for } 0 \le t \le 1. \] Then, \[ dx = dt,\quad dy = dt. \]
Step 3: Express the work done as: \[ W = \int_{0}^{1} \left[3x^2 \frac{dx}{dt} + 2y \frac{dy}{dt}\right] dt. \] Substitute \(x = 2+t\) and \(y = 1+t\): \[ W = \int_{0}^{1} \left[3(2+t)^2 + 2(1+t)\right] dt. \]
Step 4: Expand and simplify the integrand: \[ (2+t)^2 = 4 + 4t + t^2. \] Multiplying by 3: \[ 3(2+t)^2 = 12 + 12t + 3t^2. \] Also, \[ 2(1+t) = 2 + 2t. \] Thus, the integrand is: \[ 12 + 12t + 3t^2 + 2 + 2t = 14 + 14t + 3t^2. \]
Step 5: Integrate: \[ W = \int_{0}^{1} (14 + 14t + 3t^2) dt = \left[ 14t + 7t^2 + t^3 \right]_0^1. \] At \(t=1\): \[ 14(1) + 7(1)^2 + (1)^3 = 14 + 7 + 1 = 22. \] Thus, the work done, and hence the change in kinetic energy, is \(22\,\text{J}\).
Final Answer: The kinetic energy changes by \(22.0\,\text{J}\), which corresponds to option (1).
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