Chapter 6: Problem 49
A force in the \(+x\)-direction with magnitude \(F(x) = 18.0 \, \mathrm{N} - (0.530 \, \mathrm{N/m})x\) is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x = 0\), what is its speed after it has traveled 14.0 m?
Short Answer
Step by step solution
Understand the Problem
Find the Work Done by the Force
Calculate the Integral
Apply the Work-Energy Principle
Solve for the Final Velocity
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Force
- \( F(x) = 18.0 \; \mathrm{N} - (0.530 \; \mathrm{N/m})x \)
To analyze such problems, we need to integrate the force over the distance. This process will help us calculate the work done by the force, which accounts for the 'changing nature' of variable forces. This way, we accommodate for how the force's magnitude contracts gradually as the box gets further along its path.
Kinetic Energy
- The relationship between work done by a force and kinetic energy is crucial.
- Work-energy principle states: the work done on an object changes its kinetic energy.
Integrating the variable force over its path gives us the work done. The change in kinetic energy is given as, \[ W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} m(u)^2 \]Because the box starts from rest, \( u = 0 \), so the final kinetic energy equals the work done. This relationship helps us determine the box's speed after moving 14 meters.
In simpler terms, the work done by the force results in the box speeding up, and we can compute how fast by knowing the work put in and the box's mass.
Integral Calculus
- In this problem, the force changes with position, so we need to "add up" all the small contributions of force over the 14 m distance.
- This is done using an integral to summarize the continuous accumulation of force.
Once integrated, it provides the total work done on the box across the whole path from 0 to 14 meters. This result allows us to apply the work-energy principle effectively, transitioning us from mathematical calculations to physical predictions of motion outcomes.
Understanding integration in this context is pivotal because it transforms the question of "How much force is applied?" into a nuanced answer where force continuously changes, yet yields a precise work value. This can be especially insightful for students who need to tackle problems where forces are not constant.