Find the kinetic energy of a 100kg box with the x position of the box = .5t^3+2t and y=0

To find the kinetic energy of the box, we need to determine its velocity and use the kinetic energy formula:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( m = 100 \, \text{kg} \)
- \( v \) is the magnitude of the velocity vector, which we can find by differentiating the position functions with respect to time.

Given:
- \( x(t) = 0.5t^3 + 2t \)
- \( y(t) = 0 \)

### Step 1: Find the Velocity Components

1. Differentiate \( x(t) \) with respect to \( t \) to find \( v_x(t) \):
   \[
   v_x(t) = \frac{dx}{dt} = 1.5t^2 + 2
   \]

2. Differentiate \( y(t) \) with respect to \( t \) to find \( v_y(t) \):
   \[
   v_y(t) = \frac{dy}{dt} = 0
   \]

### Step 2: Calculate the Magnitude of the Velocity

Since \( y \) is constant, the velocity is only in the \( x \)-direction:
\[
v = |v_x(t)| = 1.5t^2 + 2
\]

### Step 3: Plug into the Kinetic Energy Formula

\[
KE = \frac{1}{2} \cdot 100 \cdot (1.5t^2 + 2)^2
\]

Simplify this to obtain the expression for the kinetic energy as a function of time:
\[
KE = 50 \cdot (1.5t^2 + 2)^2
\]

Would you like to evaluate this expression at a specific time or need further details?

---

1. What is the kinetic energy at \( t = 1 \) second?
2. How does the kinetic energy change with respect to time?
3. What happens to the kinetic energy if the mass is doubled?
4. How do you find the acceleration of the box from its position function?
5. How does this approach change if \( y(t) \) is not zero?

**Tip:** Kinetic energy depends on the square of velocity, so even a small increase in speed results in a larger increase in kinetic energy.

Find the kinetic energy of a 100kg box with the x position of the box = .5t^3 + 2t and y = 0.

Learn to find the kinetic energy of a 100 kg box moving along a position function x = 0.5t^3 + 2t. This solution uses calculus to derive the velocity and kinetic energy, applying core physics and kinematics concepts. Ideal for students in Grades 11-12.

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