91Ó°ÊÓ

To find the force in both the x and y directions, we can use the negative gradient of the potential energy function: \(F_x = -\frac{dU}{dx}\) and \(F_y = -\frac{dU}{dy}\) The potential energy function U is given by: \(U = -7x + 24y\) To find the force components, we will differentiate U with respect to x and y, and then find the negative of these values: \(F_x = -\frac{dU}{dx} = -\frac{-7}{dx} = 7 \mathrm{N}\) \(F_y = -\frac{dU}{dy} = -\frac{24}{dy} = -24 \mathrm{N}\)

Now we are given the mass of the particle, which is m = 5 kg. We can use Newton's second law, \(F = ma\), to calculate the acceleration components in the x and y directions: \(a_x = \frac{F_x}{m} = \frac{7}{5} \mathrm{m/s^2}\) \(a_y = \frac{F_y}{m} = \frac{-24}{5} \mathrm{m/s^2}\) Thus, the acceleration vector is \(\vec{a} = (\frac{7}{5}\hat{i} - \frac{24}{5}\hat{j}) \mathrm{m/s^2}\).

(A) We need to find the speed of the particle at t = 4 s. First, we have to find the velocity vector at 4 s and then calculate its magnitude to determine if it's 25 m/s. The velocity at t can be calculated as: \(\vec{v}(t) = \vec{u} + \vec{a}t\) At t = 4 s: \(\vec{v}(4) = (14.4\hat{i} + 4.2\hat{j}) + (\frac{7}{5}\hat{i} - \frac{24}{5}\hat{j}) \times 4\) \(\vec{v}(4) = (14.4 + \frac{28}{5})\hat{i} + (4.2 - \frac{96}{5})\hat{j}\) Speed: \(|v(4)| = \sqrt{(\vec{v}_x(4))^2 + (\vec{v}_y(4))^2} \ne 25 \mathrm{m/s}\) So, option (A) is false. (B) To check if the particle has an acceleration of 5 m/s², we can find the magnitude of the acceleration vector: \(a = |\vec{a}| = \sqrt{(a_x)^2 + (a_y)^2} \ne 5 \mathrm{m/s^2}\) So, option (B) is false. (C) To check if the acceleration is normal to the initial velocity, we can find the dot product of the velocity and acceleration vectors: \(\vec{u} \cdot \vec{a} = (14.4 \hat{i} + 4.2 \hat{j}) \cdot (\frac{7}{5}\hat{i} - \frac{24}{5}\hat{j})\) \(\vec{u} \cdot \vec{a} = 14.4 \times \frac{7}{5} + 4.2 \times \frac{-24}{5}\) \(\vec{u} \cdot \vec{a} \ne 0\) Since the dot product of the initial velocity and acceleration is not equal to 0, the acceleration is not normal to the initial velocity. So, option (C) is false. Since all options (A), (B), and (C) are false, the correct answer is (D) None of these.