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To find the force acting on the object, we'll first need to determine the acceleration of the object at the given point and then use Newton's second law, \( F = m \cdot a \).

### Given Information
- The object's path is described by the parabola \( y = 2x^2 \).
- The object is moving with a constant speed \( v = 15 \) units/sec.
- The point of interest is \( \left( 3^{1/2}, 6 \right) \).

### Step 1: Find the acceleration
Since the speed is constant, the acceleration must be centripetal and directed toward the concave side of the parabola.

#### 1.1 Parametrize the curve
The given curve is \( y = 2x^2 \). Let's parametrize it:
- \( x(t) = x \)
- \( y(t) = 2x(t)^2 \)

The vector-valued function for the curve will be:
\[
\mathbf{r}(t) = x(t) \mathbf{i} + 2[x(t)]^2 \mathbf{j}
\]
Taking derivatives with respect to \( t \):
\[
\mathbf{r}'(t) = \frac{dx}{dt} \mathbf{i} + 4x \frac{dx}{dt} \mathbf{j}
\]

Since speed \( v = 15 \), we have
\[
\sqrt{\left(\frac{dx}{dt}\right)^2 + (4x \frac{dx}{dt})^2} = 15
\]
Simplify:
\[
\frac{dx}{dt} \sqrt{1 + 16x^2} = 15
\]
At \( x = \sqrt{3} \), let's solve for \( \frac{dx}{dt} \):
\[
\frac{dx}{dt} \sqrt{1 + 16(3)} = 15
\]
\[
\frac{dx}{dt} \sqrt{49} = 15
\]
\[
\frac{dx}{dt} \cdot 7 = 15 \implies \frac{dx}{dt} = \frac{15}{7}
\]

#### 1.2 Find acceleration vector \( \mathbf{a}(t) \)
The unit tangent vector \( \mathbf{T} \) is obtained by normalizing \( \mathbf{r}'(t) \). The unit normal vector \( \mathbf{N} \) is obtained by differentiating \( \mathbf{T} \). The centripetal acceleration is \( a_c = \frac{v^2}{R} \), where \( R \) is the radius of curvature at \( x = \sqrt{3} \).

The force \( F = m a \) can then be written as \( \mathbf{F} = m \left(a_T \mathbf{T} + a_c \mathbf{N} \right) \), but with only centripetal acceleration, we calculate \( F_x \) and \( F_y \).

Let's calculate explicitly using differentiation. Let me now finalize this using computational tools to ensure accuracy.The force acting on the object due to its acceleration at the point \( \left( \sqrt{3}, 6 \right) \) is:

\[
\mathbf{F} = 0 \, \mathbf{i} + \frac{900m}{49} \, \mathbf{j}
\]

Thus, the force vector is:
\[
\mathbf{F} = \frac{900m}{49} \, \mathbf{j}
\]

This means the force has no component in the \( x \)-direction and is entirely in the \( y \)-direction.

Do you have any questions, or would you like a more detailed explanation?

### Related Questions:
1. How do we find the radius of curvature for a given curve at a point?
2. Why does the centripetal force only act perpendicular to the motion when speed is constant?
3. What would the force be if the object was accelerating tangentially as well?
4. How would the answer change if the speed was different?
5. Can this method be applied to other types of curves (e.g., cubic functions)?

### Tip:
For objects moving along curves, the force and acceleration often have significant components in the direction perpendicular to the path, known as centripetal force, especially when moving at a constant speed.

An object of mass m travels along the parabola y = 2x^2 with a constant speed of 15 units/sec. What is the force on the object due to its acceleration at (√3, 6)? (Remember Newton's law, F = ma.)

Determine the force acting on an object of mass m moving along the parabola y = 2x^2 with a constant speed of 15 units/sec, using Newton's Second Law. Includes concepts like centripetal acceleration, parametric equations, and the radius of curvature. Suitable for college-level physics and calculus students.

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