Chapter 5: Problem 36
A 25.0-kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp?
Short Answer
Step by step solution
Analyze Forces Acting on the Box
Calculate Static Friction and Normal Force
Determine the Minimum Angle for Sliding
Analyze Forces When Box is Moving
Calculate the Acceleration
Calculate Final Velocity After Sliding 5 m
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Static Friction
The formula for static friction at its maximum is given by:
- \( f_{s} = \mu_{s} N \)
- Where \( \mu_{s} = 0.35 \) in our exercise
- \( N = mg \cos \alpha \), where \( \alpha \) is the angle of the incline
Kinetic Friction
The formula for kinetic friction is similar to static friction:
- \( f_{k} = \mu_{k} N \)
- Where \( \mu_{k} = 0.25 \) in the exercise
Inclined Plane
The angle of inclination determines the plane's steepness. Its effect on motion is described by breaking down gravitational force into two components:
- Parallel to the plane (\( mg \sin \alpha \)
- Perpendicular to the plane (\( mg \cos \alpha \)
Acceleration
The net force accelerating the box downwards is the difference between the component of gravity pulling it downward and the kinetic friction opposing this pull. The formula is:
- \( a = g(\sin \alpha - \mu_{k} \cos \alpha) \)
Kinematics
For the box sliding 5 meters down the ramp, the initial velocity \( u \) is 0 since it starts from rest. Using the kinematic equation:
- \(v^2 = u^2 + 2as\)
- Where \( v \) is the final velocity
- \( a = 1.32 \, \text{m/s}^2 \)
- \( s = 5 \, \text{m} \)