Physics 201 Forces Exam Problem Example

Mastering Forces: A Complete Physics 201 Exam Problem Walkthrough

Struggling with forces problems is a universal rite of passage for any student in introductory physics. These problems form the bedrock of classical mechanics, and conquering them is non-negotiable for success in Physics 201. This article provides a complete, step-by-step deconstruction of a classic, multi-part exam problem. You won't just see the answer; you'll learn the systematic methodology to approach any forces question, transforming anxiety into confidence. We will break down a problem involving an inclined plane, friction, and Newton's laws—a staple on any midterm or final exam.

The Problem: A Classic Three-Part Challenge

Scenario: A 10.0 kg block is initially at rest on a rough inclined plane angled at 30.0° above the horizontal. A force of 50.0 N is applied parallel to the incline, directed up the slope. The coefficients of friction are μ_s = 0.50 and μ_k = 0.35. Use g = 9.80 m/s².

(a) Draw a complete free-body diagram (FBD) for the block, labeling all forces with their symbolic names and, where possible, their magnitudes. (b) Determine the magnitude of the block's acceleration immediately after it starts moving. (c) Calculate the minimum magnitude of an additional force (applied parallel to the incline, but down the slope) that must be applied to keep the block moving up the incline at a constant velocity.

Part 1: The Foundation – Crafting the Perfect Free-Body Diagram

Before a single calculation, you must visualize the physics. The free-body diagram (FBD) is your most powerful tool. It isolates the object (the block) and represents all external forces acting upon it as vectors.

Step 1: Coordinate System. Always align your coordinate axes with the incline. This is the golden rule. Set the +x-axis parallel to the slope, pointing up the incline. Set the +y-axis perpendicular to the slope, pointing away from the plane. This choice simplifies math because the normal force and friction will align with your axes.

Step 2: Identify and Draw Forces.

1. Gravity (Weight): F_g = m*g acts vertically downward. In our tilted coordinates, it has two components:
   • Parallel component: F_g,x = m*g*sin(θ) (down the slope, so negative in our +x system).
   • Perpendicular component: F_g,y = m*g*cos(θ) (into the plane, negative in our +y system).
2. Normal Force (N): The force exerted by the incline perpendicular to its surface. It acts in the +y direction. Its magnitude is not simply m*g because of the incline.
3. Applied Force (F_app): Given as 50.0 N up the slope. This is in our +x direction.
4. Friction (f): Since the block will move (we'll confirm this), we use kinetic friction. It always opposes relative motion. The block is trying to move up the slope due to F_app, so friction acts down the slope, in the -x direction. Its magnitude is f_k = μ_k * N.

Your FBD should show these four vectors originating from the block's center, with clear labels (F_g, N, F_app, f_k). For part (a), you would calculate and write the numerical values for F_g (98.0 N), F_g,x (49.0 N), and F_g,y (84.9 N). The normal force N and friction f_k depend on N, so you leave them as symbols or calculate N first.

Part 2: The Motion – Calculating Acceleration (Part b)

The block starts from rest. We must first check if the applied force overcomes static friction to initiate motion.

Check for Motion:

• Forces parallel to incline (x-direction) before motion: F_app (up) vs. F_g,x (down) + f_s,max (down).
• f_s,max = μ_s * N.
• First, find N from the y-direction equilibrium (no acceleration perpendicular to the plane): ΣF_y = 0.
  • N - F_g,y = 0 → N = F_g,y = m*g*cos(θ) = 10.0 * 9.80 * cos(30°) = 84.9 N.
• f_s,max = 0.50 * 84.9 N = 42.45 N.
• Total force trying to hold it back: F_g,x + f_s,max = 49.0 N + 42.45 N = 91.45 N.
• Applied force is only 50.0 N. Conclusion: The block will not move up. It will either stay at rest or slide down.

Wait! This is a critical exam insight. The problem states "immediately after it starts moving." This implies we must first find the condition for it to start moving at all. Our calculation shows 50 N up is insufficient to overcome the maximum static friction plus the downhill gravity component. Therefore, the block will not move up; it will remain at rest unless a larger force is applied.

Re-reading the problem: "A force of 50.0 N is applied parallel to the incline, directed up the slope."