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 \( K \) of the box, we need to determine its velocity. The kinetic energy of an object with mass \( m \) and velocity \( v \) is given by:

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

Given:
- Mass, \( m = 100 \, \text{kg} \)
- Position function in the \( x \)-direction: \( x(t) = 0.5t^3 + 2t \)
- Position function in the \( y \)-direction: \( y(t) = 0 \)

Since \( y(t) = 0 \), the motion is only in the \( x \)-direction, so the velocity will be determined solely by \( x(t) \).

### Step 1: Find the velocity in the \( x \)-direction
The velocity in the \( x \)-direction, \( v_x(t) \), is the derivative of \( x(t) \) with respect to time \( t \):

\[
v_x(t) = \frac{dx}{dt} = \frac{d}{dt} \left(0.5t^3 + 2t\right)
\]

Calculating the derivative:

\[
v_x(t) = 1.5t^2 + 2
\]

### Step 2: Calculate the magnitude of velocity \( v(t) \)
Since there is no \( y \)-component of velocity, the total velocity \( v(t) \) is just \( v_x(t) \):

\[
v(t) = v_x(t) = 1.5t^2 + 2
\]

### Step 3: Substitute into the kinetic energy formula
Now, substitute \( m = 100 \) kg and \( v(t) = 1.5t^2 + 2 \) into the kinetic energy formula:

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

Simplifying:

\[
K(t) = 50 \times (1.5t^2 + 2)^2
\]

This expression gives the kinetic energy as a function of time \( t \).

Would you like me to evaluate this for a specific time or delve into further details?

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**Follow-up Questions:**

1. How would the kinetic energy change if the mass of the box increased?
2. What happens to the kinetic energy if the \( y \)-position is also time-dependent?
3. How would you calculate the velocity if the box moved in both \( x \)- and \( y \)-directions?
4. What effect does the term \( 0.5t^3 \) in \( x(t) \) have on the acceleration of the box?
5. How does the power (rate of energy change) vary with respect to \( t \)?

**Tip:** To analyze the kinetic energy at a specific instant, simply substitute the desired \( t \)-value into the velocity function before plugging into the kinetic energy formula.

Determine the kinetic energy of a 100 kg box moving along the x-axis with position given by x(t) = 0.5t^3 + 2t. This problem covers concepts of kinetic energy and differentiation, providing a step-by-step solution to find velocity and kinetic energy. Suitable for students in Grades 11-12.

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