EX. 3: A 250 kg box slides down a ramp and then across a level floor The coefficient of kinetic friction along the floor is 0.20. A person sees the a box moving at 1.0m/s[left] and pushes on it with a horizontal force of 140 N [right]. a. How far does the box travel before coming to rest? b. How will the results change if the box is moving right and the person still pushes right with the same force?

**a. Box moving left, force applied to the right:**<br /><br />1. **Free Body Diagram:** Draw a free-body diagram of the box. Forces acting on it are:<br /> * Weight (mg) downwards<br /> * Normal force (N) upwards<br /> * Kinetic friction (f_k) to the right (opposing motion)<br /> * Applied force (F_app) to the right<br /><br />2. **Net Force:** The net force in the horizontal direction is the vector sum of the applied force and the friction force. Since they act in opposite directions:<br /> F_net = F_app - f_k<br /><br />3. **Friction Force:** Calculate the friction force:<br /> f_k = μ_kN <br /> Since the box is on a level floor, N = mg. Therefore:<br /> f_k = μ_kmg = (0.20)(250 kg)(9.8 m/s²) = 490 N<br /><br />4. **Net Force Calculation:**<br /> F_net = 140 N - 490 N = -350 N (The negative sign indicates the net force is to the left, opposing the initial motion).<br /><br />5. **Acceleration:** Use Newton's second law (F_net = ma) to find the acceleration:<br /> a = F_net / m = -350 N / 250 kg = -1.4 m/s² (Deceleration)<br /><br />6. **Distance Traveled:** Use the following kinematic equation, where v_f is the final velocity (0 m/s), v_i is the initial velocity (-1.0 m/s - negative because it's to the left), and a is the acceleration:<br /> v_f² = v_i² + 2ad<br /> 0 = (-1.0 m/s)² + 2(-1.4 m/s²)d<br /> d = (1.0 m²/s²) / (2.8 m/s²) <br /> d ≈ 0.36 m<br /><br />**b. Box moving right, force applied to the right:**<br /><br />1. **Free Body Diagram:** The free-body diagram is the same as before, except the friction force now acts to the *left* (opposing the rightward motion).<br /><br />2. **Net Force:**<br /> F_net = F_app - f_k = 140 N - 490 N = -350 N (Again, negative, indicating a deceleration)<br /><br />3. **The rest of the calculations are identical to part (a).** The box will travel approximately 0.36 m before coming to rest.<br /><br />**Conclusion:**<br /><br />In both scenarios, the box travels the same distance before stopping. Even though the initial direction of motion is different, the force of friction and the applied force create the same net force (in magnitude) and therefore the same deceleration.<br />