You push a box with a force given by the function $F\left(x\right)=13-0.6x$. Assuming there's no friction forces, and the box is initially at rest at $x=0$, how fast is the box moving after it has traveled a distance of $20m$?

• A. $5.37\text{ m/s}$
• B. $2.83\text{ m/s}$
• C. $7.48\text{ m/s}$
• D. $24.95\text{ m/s}$

Answer: (C)

Verified step by step guidance

1. Identify the given information: mass \(m = 5\operatorname{\mathrm{kg}}\), force as a function of position \(F(x) = 13 - 0.6x\), initial position \(x=0\), and displacement \(\Delta x = 20\,\mathrm{m}\). The box starts from rest, so initial velocity \(v_0 = 0\).
2. Recall that work done by a variable force over a displacement from \(x=0\) to \(x=20\) is given by the integral of the force over that distance: \(W = \int_0^{20} F(x) \, dx\).
3. Calculate the work done by integrating the force function: \(W = \int_0^{20} (13 - 0.6x) \, dx\). This will give the total work done on the box as it moves 20 meters.
4. Use the work-energy theorem, which states that the net work done on an object equals its change in kinetic energy: \(W = \Delta K = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2\). Since the box starts from rest, \(v_0 = 0\), so \(W = \frac{1}{2} m v^2\).
5. Solve the equation \(W = \frac{1}{2} m v^2\) for the final velocity \(v\): \(v = \sqrt{\frac{2W}{m}}\). Substitute the value of \(W\) from the integral and the mass \(m\) to find the speed after traveling 20 meters.