Question

A particle experiences a variable force F=(4x\(\hat i\)+3y²\(\hat j\)) in a horizontal x–y plane. Assume distance in meters and force is Newton.If the particle moves from point (1, 2) to point (2, 3) in the x–y plane; then Kinetic Energy changes by

• A. 50.0 J
• B. 12.5 J
• C. 25.0 J
• D. 0 J

Answer 1

The Correct Option is C

To solve the problem of determining the change in kinetic energy when a particle moves under a variable force from one point to another, we need to calculate the work done by the force along the path from the initial to the final position. The change in kinetic energy is equal to the work done on the particle.

The force is given by F = (4x\hat{i} + 3y^2\hat{j}) \, \text{N}. The particle moves from point \( A(1, 2) \) to point \( B(2, 3) \).

The work done by a force F = F_x\hat{i} + F_y\hat{j} along a path from point \(A\) to point \(B\) is given by:

W = \int_{\text{path}} \mathbf{F} \cdot d\mathbf{r}

Since the force is conservative, we can express this as:

W = \int_{x_1}^{x_2} F_x \, dx + \int_{y_1}^{y_2} F_y \, dy

Substituting the given limits and force components, we evaluate:

• W_x = \int_{1}^{2} 4x \, dx = 4 \left[ \frac{x^2}{2} \right]_{1}^{2} = 4 \left(\frac{4}{2} - \frac{1}{2} \right) = 4 \cdot \frac{3}{2} = 6 \, \text{J}
• W_y = \int_{2}^{3} 3y^2 \, dy = 3 \left[ \frac{y^3}{3} \right]_{2}^{3} = \left(27 - 8 \right) = 19 \, \text{J}

Thus, the total work done, and therefore the change in kinetic energy, is:

W = W_x + W_y = 6 + 19 = 25 \, \text{J}

Hence, the kinetic energy changes by 25.0 J as the particle moves from point (1, 2) to point (2, 3) in the x-y plane.

The correct answer is: 25.0 J