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.