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Pushing a Cat. Your cat "Ms" (mass 7.00 \(\mathrm{kg}\) ) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by excring a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

Short Answer

Expert verified
Ms's speed at the top of the ramp is approximately 6.58 m/s.

Step by step solution

01

Understand the Given Problem

We need to determine the speed of a cat named Ms, who initially moves at 2.40 m/s up a frictionless incline that is 2 m long, with a push of 100 N along the ramp. The incline is at a 30° angle. We'll apply the work-energy theorem to solve this problem.
02

Identify the Variables and Equations

Using the work-energy theorem, the work done on the cat equals the change in kinetic energy: \[ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]Where:- \( m = 7.00 \text{ kg} \) (mass of the cat)- \( v_i = 2.40 \text{ m/s} \) (initial velocity)- \( L = 2.00 \text{ m} \) (length of the ramp)- \( F = 100 \text{ N} \) (force exerted parallel to the ramp)Also, the gravitational force acting down the ramp does work against the applied force.
03

Calculate the Work Done by the Applied Force

The work done by the applied force is given by:\[ W_{\text{applied}} = F \cdot L \]Substitute the known values:\[ W_{\text{applied}} = 100 \text{ N} \times 2.00 \text{ m} = 200 \text{ J} \]
04

Calculate the Work Done by Gravity

The gravitational force does negative work along the ramp, given by:\[ W_{\text{gravity}} = -m g L \sin(\theta) \]Substitute the known values:\[ W_{\text{gravity}} = -(7.00 \text{ kg}) \times 9.81 \text{ m/s}^2 \times 2.00 \text{ m} \times \sin(30.0^\circ) = -68.67 \text{ J} \]
05

Use the Work-Energy Theorem to Find Final Speed

Using the work-energy theorem:\[ W_{\text{net}} = W_{\text{applied}} + W_{\text{gravity}} = \Delta KE \]\[ 200 \text{ J} - 68.67 \text{ J} = \frac{1}{2} \times 7.00 \text{ kg} \times v_f^2 - \frac{1}{2} \times 7.00 \text{ kg} \times (2.40 \text{ m/s})^2 \]\[ 131.33 \text{ J} = \frac{1}{2} \times 7.00 \text{ kg} \times v_f^2 - 20.16 \text{ J} \]\[ 151.49 \text{ J} = \frac{1}{2} \times 7.00 \text{ kg} \times v_f^2 \]\[ v_f^2 = \frac{151.49 \text{ J}}{3.5 \text{ kg}} \]\[ v_f = \sqrt{43.28 \text{ m}^2/\text{s}^2} \approx 6.58 \text{ m/s} \]
06

Verify Calculations and Conclusion

After solving for \( v_f \), we find Ms's final speed to be approximately 6.58 m/s at the top of the ramp. This approach ensures that all forces and the initial motion are accounted for.

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Key Concepts

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

Kinetic Energy and the Work-Energy Theorem
Kinetic energy is a measure of the energy an object has due to its motion. It is given by the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In the exercise involving Ms the cat, her initial kinetic energy is calculated using her initial velocity of 2.40 m/s. This value represents how much energy she has as she starts moving up the incline.

The work-energy theorem states that the work done on an object equals the change in its kinetic energy. In simpler terms, if you do work on an object, you are either increasing or decreasing its kinetic energy, depending on the direction of the force applied. For Ms the cat, the net work done (the difference between applied work and gravitational work) increases her kinetic energy, allowing her to gain speed as she moves up the ramp.

By applying the work-energy theorem, we add the work done by our push and subtract the work done by gravity to find out Ms's final speed. This method effectively gives insight into how energy transformations take place as Ms moves along the frictionless surface.
Understanding Gravitational Force on an Incline
Gravitational force is what pulls objects towards the Earth's surface, contributing to weight. When an object is on an incline, gravity still acts downward, but it affects the motion along the ramp. Importantly, gravitational force can be broken down into components: one perpendicular to the incline and one parallel to it.

In this exercise, gravitational force acts against the cat’s motion up the incline. The component of gravitational force along the ramp is calculated using \( W_{\text{gravity}} = -m g L \sin(\theta) \), where \( \theta \) is the angle of the incline, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( L \) is the length of the incline. This formula helps determine how much gravitational force slows down the motion of the cat.

The negative sign indicates that gravity does negative work—it reduces the kinetic energy of Ms the cat. Understanding this concept is crucial, as it helps in calculating the net work done on an object, offering clarity on how gravitational and other forces interact on inclined planes.
Frictionless Incline and Its Implications
A frictionless incline is an idealized situation where there is no friction between an object and the surface. This means that no energy is lost to heat or resistance as the object moves. In the problem at hand, the frictionless incline allows force calculations to focus solely on the applied and gravitational forces without worrying about friction.

This freedom from friction means every bit of energy from the push goes into overcoming gravitational force and increasing the cat's kinetic energy. It simplifies calculations, as one does not have to subtract any energy lost to friction, which can be complex and variable depending on the surface. For students, understanding a frictionless incline serves as an excellent introduction to basic physics principles. It allows them to recognize and compute energy transformations without additional factors complicating the scenario. Hence, it provides a clear picture of how forces and energy work in a relatively simplified system, a fundamental understanding before approaching more challenging problems that include friction.

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