An object of mass m travels along the parabola 

yequals=x squaredx2 

with a constant speed of 

1414 

​units/sec. What is the force on the object due to its acceleration at 

left parenthesis 6 Superscript 1 divided by 2 Baseline comma 6 right parenthesis61/2,6​? 

​(Remember Newton's​ law, 

**F**equals=m**a**​.)

Question content area bottom

Part 1

**F**equals=left parenthesis nothing right parenthesisenter your response here 

**i**plus+left parenthesis nothing right parenthesisenter your response here 

**j**

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

To find the force on the object due to its acceleration at the given point, follow these steps:

1. **Parameterize the motion along the parabola.**

   Given:
   - The object's path is \( y = x^2 \).
   - The constant speed is \( 14 \) units/sec.
   - We need to find the force at the point \( \left( \sqrt{6}, 6 \right) \).

2. **Parameterize the coordinates using \( t \).**

   Let \( x = t \), then \( y = t^2 \).
   
   The position vector \( \mathbf{r}(t) \) is:
   \[
   \mathbf{r}(t) = (t, t^2)
   \]

3. **Find the velocity components.**

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

   The magnitude of the velocity \( v \) is given as \( 14 \):
   \[
   v = \sqrt{v_x^2 + v_y^2} = 14
   \]
   \[
   \sqrt{1 + (2t)^2} = 14
   \]
   \[
   1 + 4t^2 = 196
   \]
   \[
   4t^2 = 195
   \]
   \[
   t^2 = \frac{195}{4} = 48.75
   \]
   \[
   t = \sqrt{48.75} = 3\sqrt{3}
   \]

4. **Find the acceleration components.**

   The acceleration components are the time derivatives of the velocity components:
   \[
   a_x = \frac{dv_x}{dt} = 0
   \]
   \[
   a_y = \frac{dv_y}{dt} = \frac{d(2t)}{dt} = 2
   \]

   Therefore, the acceleration vector is:
   \[
   \mathbf{a} = 0 \mathbf{i} + 2 \mathbf{j}
   \]

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

   \[
   \mathbf{F} = m (0 \mathbf{i} + 2 \mathbf{j}) = 0 \mathbf{i} + 2m \mathbf{j}
   \]

   Hence, the force on the object is:
   \[
   \mathbf{F} = \left( 0 \right) \mathbf{i} + \left( 2m \right) \mathbf{j}
   \]

So, the force on the object due to its acceleration at \( \left( \sqrt{6}, 6 \right) \) is:
\[
\mathbf{F} = \left( 0 \right) \mathbf{i} + \left( 2m \right) \mathbf{j}
\]

Do you want details or have any questions?

Here are some questions you might consider next:
1. How do you calculate the centripetal force for an object moving in a circular path?
2. What is the impulse imparted to an object during a collision?
3. How do you determine the gravitational force between two masses?
4. What is the difference between conservative and non-conservative forces?
5. How do you find the torque acting on a rotating object?

**Tip:** When working with parametric equations, always make sure to correctly differentiate with respect to the parameter and apply the chain rule if necessary.

Learn how to calculate the force on an object moving along the parabola y = x^2 at a specific point (sqrt(6), 6). This problem involves parametric equations, velocity and acceleration vectors, and Newton's second law of motion (F = ma). Suitable for high school students studying physics and calculus.

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