14. A horizontal force of magnitude $$ F=10 \mathrm{~N} $$ pushes a $$ 4.0-\mathrm{kg} $$ box initially at rest along a frictionless floor. What is the box's speed (v) after traveling a distance of $$ \Delta x=5.0 \mathrm{~m} $$ ? A. $$ 4.0 \mathrm{~m} / \mathrm{s} $$ B. $$ 1.0 \mathrm{~m} / \mathrm{s} $$ C. $$ 3.0 \mathrm{~m} / \mathrm{s} $$ D. $$ 2.0 \mathrm{~m} / \mathrm{s} $$ E. $$ 5.0 \mathrm{~m} / \mathrm{s} $$

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UpStudy AI · Accepted Answer

Given:
- Force: $$ F = 10\,\mathrm{N} $$
- Mass: $$ m = 4.0\,\mathrm{kg} $$
- Displacement: $$ \Delta x = 5.0\,\mathrm{m} $$
- Initial velocity: $$ v_0 = 0\,\mathrm{m/s} $$

Since the floor is frictionless, the work done by the force is fully converted into kinetic energy.

1. **Calculate the work done $$ W $$:**

$$
   W = F \Delta x = 10\,\mathrm{N} \times 5.0\,\mathrm{m} = 50\,\mathrm{J}
   $$

2. **Set the work equal to the kinetic energy (KE):**

$$
   KE = \frac{1}{2} m v^2 = 50\,\mathrm{J}
   $$

3. **Solve for the speed $$ v $$:**

$$
   \frac{1}{2} (4.0\,\mathrm{kg}) v^2 = 50\,\mathrm{J}
   $$

Multiply both sides by 2:

$$
   (4.0\,\mathrm{kg}) v^2 = 100\,\mathrm{J}
   $$

Divide both sides by $$ 4.0\,\mathrm{kg} $$:

$$
   v^2 = \frac{100\,\mathrm{J}}{4.0\,\mathrm{kg}} = 25\,\mathrm{m^2/s^2}
   $$

Take the square root:

$$
   v = \sqrt{25\,\mathrm{m^2/s^2}} = 5.0\,\mathrm{m/s}
   $$

Thus, the box's speed after traveling $$ 5.0\,\mathrm{m} $$ is:

E. $$ 5.0\,\mathrm{m/s} $$
The box's speed after traveling 5.0 meters is 5.0 m/s, so the correct answer is E.

A horizontal force of magnitude pushes a box initially at rest along a
frictionless floor. What is the box’s speed (v) after traveling a distance of ?
A.
B.
C.
D.
E.

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A horizontal force of magnitude pushes a box initially at rest along a frictionless floor. What is the box’s speed (v) after traveling a distance of ? A. B. C. D. E.