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A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the \(+x\) axis. As a result, the kinetic energy of the sled increases by \(38 \% .\) By what percentage would the sled's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

Short Answer

Expert verified
The sled's kinetic energy would increase by approximately 17.8%.

Step by step solution

01

Understanding the Problem

The sled is initially being pulled in the same direction as its displacement along the \(+x\) axis, resulting in a \(38\%\) increase in kinetic energy. We need to find the percentage increase in kinetic energy when the applied force makes a \(62^{\circ}\) angle with the \(+x\) axis.
02

The Effect of the Force Angle on Work Done

Work done by a force is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force magnitude, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement directions. Initially, \( \theta = 0^{\circ} \) and \( \cos(0^{\circ}) = 1 \). At \( 62^{\circ} \), \( \cos(62^{\circ}) \) has a value less than 1, affecting the work done.
03

Calculating the Ratio of work done

The ratio of work done with different angles is \( \frac{W_{new}}{W_{original}} = \frac{F \cdot d \cdot \cos(62^{\circ})}{F \cdot d}\). Simplifying gives \( \cos(62^{\circ}) \).
04

Computing \(\cos(62^{\circ})\)

Determine \( \cos(62^{\circ}) \) using a calculator, which is approximately \( 0.4695 \). This is the factor that modifies the work done, and consequently, the change in kinetic energy.
05

Calculate the New Percentage Increase in Kinetic Energy

If the full force along the \(+x\) axis led to a \(38\%\) increase, the new increase percentage is \( 0.4695 \times 38\% \). Compute this as \( 0.4695 \times 38 = 17.841\% \).
06

Final Step: Conclusion

The kinetic energy increase of the sled would be approximately \( 17.8\% \) when pulled at a \(62^{\circ}\) angle to the \(+x\) axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Force
When a sled is being pulled across snow, the work done by the applied force fundamentally changes the sled's kinetic energy. Work done is a measure of energy transfer that occurs when an object is moved by a force. The formula for calculating work done is given by \( W = F \cdot d \cdot \cos(\theta) \). Here, \( F \) is the magnitude of the force applied, \( d \) is the displacement of the sled, and \( \theta \) is the angle between the direction of the force and the direction of displacement.

In our scenario, when the force is aligned perfectly with the displacement (angle zero), the work done achieves its maximum value, solely determined by the force and the distance. This situation illustrates why the sled experiences a 38% increase in kinetic energy initially. However, if the angle between them changes, it impacts the effectiveness of the force in doing work on the sled.
Angle of Force
The angle of force is a crucial element that affects the amount of work that can be done on an object.

- When the force is perfectly aligned with the displacement (\( \theta = 0^{\circ} \)), the full force contributes to the movement since \( \cos(0^{\circ}) = 1 \).
- However, as the angle increases, like in our problem where \( \theta = 62^{\circ} \), the component of the force contributing to the motion decreases.
- This is because the cosine of the angle, \( \cos(\theta) \), decreases, meaning less work is performed in the direction of motion.

In this problem, the angle change reduces the effective work by about half when considering the new direction of 62 degrees.
Cosine Function
The cosine function is vital in physics for calculating work done, particularly when dealing with angles between force and displacement.

- The cosine of an angle in a right triangle corresponds to the adjacent side divided by the hypotenuse.
- It determines how much of the force actually translates into movement across a particular direction.
- In our problem, the angle \( \theta \) changes to 62 degrees, making \( \cos(62^{\circ}) \) crucial for reevaluating the impact of the force.
- We find that \( \cos(62^{\circ}) \approx 0.4695 \), indicating less than half of the force promotes the sled's horizontal movement.

This reduced cosine value results in a proportionally smaller increase in the sled's kinetic energy.
Percentage Increase
Understanding the percentage increase in kinetic energy involves comparing two scenarios: one where the force is directly aligned with the motion and another where the angle differs.

- Initially, with direct application, the kinetic energy increases by 38%.
- With the force applied at an angle (62 degrees in our case), the effectiveness of the force diminishes according to the cosine factor.
- We calculate the new percentage increase by multiplying the cosine of the angle with the initial percentage increase. So, the new increase is \( 0.4695 \times 38\% = 17.841\% \).

The decrease from 38% to roughly 17.8% highlights the significant influence of force direction on the sled's acceleration.
Displacement Along Axis
Displacement along the axis refers to how far the sled moves in a particular direction, here represented as the \(+x\) axis. In physics, displacement is not just distance; it's a vector, meaning it has both magnitude and direction.

- Initially, the force and displacement are along the same axis, maximizing the work done by the force and leading to the full potential increase in kinetic energy.
- A change in the force's direction, as when it points 62 degrees above the \(+x\) axis, affects not only how much work is done but also how effectively the sled moves forward along this axis.
- The cosine function plays a role in calculating how much of the force's work still contributes to this horizontal displacement.

Thus, both the direction and magnitude of displacement are key to understanding motion dynamics in physics.

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Most popular questions from this chapter

A person pulls a toboggan for a distance of \(35.0 \mathrm{~m}\) along the snow with a rope directed \(25.0^{\circ}\) above the snow. The tension in the rope is \(94.0 \mathrm{~N}\). (a) How much work is done on the toboggan by the tension force? (b) How much work is done if the same tension is directed parallel to the snow?

A \(55-\mathrm{kg}\) box is being pushed a distance of \(7.0 \mathrm{~m}\) across the floor by a force \(\overrightarrow{\mathrm{P}}\) whose magnitude is \(150 \mathrm{~N}\). The force \(\overrightarrow{\mathrm{P}}\) is parallel to the displacement of the box. The coefficient of kinetic friction is \(0.25 .\) Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

The (nonconservative) force propelling \(1.50 \times 10^{3}-\mathrm{kg}\) a car up a mountain road does \(4.70 \times 10^{6} \mathrm{~J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{~m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. (a) What type of energy is changing? (b) Is the work being done by the net external force acting on the skier positive, zero, or negative? Why? (c) How is this work related to the change in the energy of the skier? A 70.3 -kg water-skier has an initial speed of \(6.10 \mathrm{~m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{~m} / \mathrm{s}\). Determine the work done by the net external force acting on the skier.

A car accelerates uniformly from rest to \(20.0 \mathrm{~m} / \mathrm{s}\) in \(5.6 \mathrm{~s}\) along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is \(9.0 \times 10^{3} \mathrm{~N},\) and \((\mathrm{b})\) the weight of the car is \(1.4 \times 10^{4} \mathrm{~N}\)

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