An object of mass m travels along the parabola yequals6 x squared with a constant speed of 8 ​units/sec. What is the force on the object due to its acceleration at left parenthesis 2 Superscript 1 divided by 2 Baseline comma 12 right parenthesis​? ​(Remember Newton's​ law, Fequalsma​.)
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Part 1
Fequalsleft parenthesis nothing right parenthesis

iplusleft parenthesis nothing right parenthesis

j
​(Type exact​ answers, using radicals as needed. Type expressions using m as the​ variable.)

To solve for the force on the object due to its acceleration, we need to follow these steps:

1. **Determine the acceleration of the object.**

   Given:
   - The object's path is \( y = 6x^2 \).
   - The speed is constant at 8 units/sec.
   - We need to find the force at the point \( \left( \sqrt{2}, 12 \right) \).

2. **Find the velocity components.**

   Since the speed is constant, we use the relationship between speed and velocity:
   \[
   v = \sqrt{v_x^2 + v_y^2} = 8 \text{ units/sec}
   \]

   To find \( v_x \) and \( v_y \), we can parameterize the curve \( y = 6x^2 \).

   Let's parameterize \( x \) as \( x = t \):
   \[
   y = 6t^2
   \]

   The parameterization of the position is \( \mathbf{r}(t) = (t, 6t^2) \).

   The velocity components are the time derivatives of the position components:
   \[
   v_x = \frac{dx}{dt} = 1
   \]
   \[
   v_y = \frac{dy}{dt} = \frac{d(6t^2)}{dt} = 12t
   \]

   Using \( v = \sqrt{v_x^2 + v_y^2} = 8 \):
   \[
   \sqrt{1 + (12t)^2} = 8
   \]
   \[
   1 + 144t^2 = 64
   \]
   \[
   144t^2 = 63
   \]
   \[
   t^2 = \frac{63}{144} = \frac{7}{16}
   \]
   \[
   t = \pm \frac{\sqrt{7}}{4}
   \]

   We will use the positive value:
   \[
   t = \frac{\sqrt{7}}{4}
   \]

3. **Calculate the acceleration components.**

   The acceleration components are the time derivatives of the velocity components:
   \[
   a_x = \frac{dv_x}{dt} = 0
   \]
   \[
   a_y = \frac{dv_y}{dt} = \frac{d(12t)}{dt} = 12
   \]

   Therefore, the acceleration vector is:
   \[
   \mathbf{a} = 0 \mathbf{i} + 12 \mathbf{j}
   \]

4. **Apply Newton's second law \( \mathbf{F} = m \mathbf{a} \).**

   \[
   \mathbf{F} = m (0 \mathbf{i} + 12 \mathbf{j}) = 0 \mathbf{i} + 12m \mathbf{j}
   \]

   Hence, the force on the object is:
   \[
   \mathbf{F} = 0 \mathbf{i} + 12m \mathbf{j}
   \]

So, the force on the object due to its acceleration at \( \left( \sqrt{2}, 12 \right) \) is:
\[
\mathbf{F} = \left( 0 \right) \mathbf{i} + \left( 12m \right) \mathbf{j}
\]

Do you want details or have any questions?

Here are some questions you might consider next:
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**Tip:** Always remember to check the units of your final answer to ensure they are consistent with the physical quantities involved.

Learn how to calculate the force on an object traveling along a parabolic path using Newton's second law of motion. Detailed step-by-step solution for finding force due to acceleration at a specific point on the path. Suitable for advanced high school students and physics enthusiasts.

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