Question

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

The speed of the box is approximately 6.06 m/s.

Step-by-step solution

Understand the Problem

We're given a force function that varies with position, \( F(x) = 18.0 \, \mathrm{N} - (0.530 \, \mathrm{N/m})x \). We need to find the speed of a 6.00-kg box after it travels 14.0 m starting from rest.

Find the Work Done by the Force

The work done by a variable force as an object moves from \( x = 0 \) to \( x = 14 \, \mathrm{m} \) can be calculated by integrating the force function:\[W = \int_{0}^{14} F(x) \, dx = \int_{0}^{14} \left( 18.0 - 0.530x \right) \, dx\]Calculate the integral to find the work done.

Calculate the Integral

Solve:\[W = \int_{0}^{14} (18.0 - 0.530x) \, dx \]This evaluates to:\[W = \left[ 18.0x - 0.530 \frac{x^2}{2} \right]_{0}^{14} = (18.0 \times 14) - 0.530 \times 7 \times (14)^2 \]Calculate this to find \( W = 110.0 \, \mathrm{J} \).

Apply the Work-Energy Principle

According to the work-energy principle, the work done on the box is equal to the change in kinetic energy:\[W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} m(0)^2 \]Substitute the work and solve for velocity \( v \).

Solve for the Final Velocity

We know:\[110.0 = \frac{1}{2} \times 6.0 \times v^2\]Rearrange to solve for \( v \):\[v^2 = \frac{110.0 \times 2}{6.0} = 36.67 \]\[v = \sqrt{36.67} \approx 6.06 \, \mathrm{m/s}\]

Conclusion

The speed of the box after it has traveled 14.0 m is approximately \( 6.06 \, \mathrm{m/s} \).

Key concepts

Variable Force

In physics, when we talk about variable force, we refer to a force whose magnitude changes depending on certain factors, such as position, time, or velocity of the object it is acting upon. In the given exercise, the force is expressed as a function of position:

• \( F(x) = 18.0 \; \mathrm{N} - (0.530 \; \mathrm{N/m})x \)

This equation means the force decreases as the position \( x \) increases. The more to the right the box moves, the lesser the force applied forward. This could be due to a medium's resistance, a sort of friction, or adjusting thrust.
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

Kinetic energy is the energy of motion. When an object moves, it has kinetic energy. Our exercise uses this idea prominently.

• 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.

For our box on a frictionless surface, the work done by the variable force as it moves from \( x = 0 \) to \( x = 14 \; \mathrm{m} \) is calculated using this principle.
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

Integral calculus is a branch of mathematics that deals with the concept of integration which allows us to determine quantities like area, volume, and in our case, work done by a force.

• 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.

The integral we compute is \[ W = \int_{0}^{14} (18.0 - 0.530x) \; dx \]This integral splits the problem into small segments over which we calculate force exerted and multiplied by the distance traveled.
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.