/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A cookie jar is moving up a \(40... [FREE SOLUTION] | 91Ó°ÊÓ

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A cookie jar is moving up a \(40^{\circ}\) incline. At a point \(55 \mathrm{~cm}\) from the bottom of the incline (measured along the incline), the jar has a speed of \(1.4 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between jar and incline is \(0.15 .\) (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?

Short Answer

Expert verified
(a) Determine \( d \) using conservation of energy. (b) Solve for final velocity using energy equations. (c) Both answers increase with less friction.

Step by step solution

01

Identify the forces involved

The forces acting on the jar are gravity, normal force, and friction. The gravitational force component parallel to the incline is \( mg \sin \theta \), and the component perpendicular to the incline is \( mg \cos \theta \). The frictional force, which opposes motion, is calculated by \( f_k = \mu_k N \), where \( N = mg \cos \theta \).
02

Solve for distance jar will slide up the incline

Using energy conservation, the initial kinetic energy (KE) plus work done against friction equals the potential energy (PE) at the highest point. The equation \( \frac{1}{2}mv^2 - f_k d = mgh \) can be set up, where \( d \) is the distance traveled up the incline, and \( h = (d+0.55)\sin{40^\circ} \). Solving for \( d \), identify that friction does work against the kinetic energy hence reducing the distance traveled.
03

Calculate the speed at the bottom after sliding back

When the jar slides back to the bottom, energy conservation points that the initial potential energy at the top converts back to kinetic energy, minus the work done by friction. Use the equation \( mgh - f_k (d + 0.55) = \frac{1}{2} m v^2 \) to solve for \( v \), the final speed when it returns to the bottom.
04

Analyze the impact of decreasing the coefficient of friction

A decrease in the coefficient of kinetic friction implies less friction opposing the motion. Thus, for (a), it implies the jar will potentially move further up as there is less resistance. For (b), decreasing the friction means more kinetic energy remains as it returns down the incline, so the speed at the bottom might be greater.

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Key Concepts

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

Coefficient of Friction
The coefficient of friction is a measure of how much frictional force exists between two surfaces in contact. It's a crucial concept in understanding motion on an inclined plane. In this exercise, the coefficient of kinetic friction is given as 0.15.
This means that 15% of the force perpendicular to the surface resists motion. Friction is always in the direction to oppose motion. It's calculated using the formula \[f_k = \mu_k N \]where:
  • \( f_k \) is the kinetic frictional force
  • \( \mu_k \) is the coefficient of kinetic friction
  • \( N \) is the normal force
The normal force is usually the component of the object's weight that presses perpendicularly against the surface.Understanding this helps us predict how the jar will behave on the surface. If you reduce the coefficient of friction, the frictional force decreases, allowing the jar to travel farther up the incline.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, less than 90°. It allows an object to be moved upwards or downwards while overcoming gravitational forces more gradually than lifting directly vertical. In this exercise:
  • The incline is tilted at an angle of \( 40^\circ \)
  • The initial speed is \( 1.4 \, \text{m/s} \)
  • The jar starts \( 0.55 \text{ m} \) away from the bottom
To analyze the motion on an inclined plane, we break down forces into components:- Parallel to the incline: gravity pushes the object down.- Perpendicular to the incline: gravity is countered by the normal force.The formula for the gravitational force component parallel to the incline is \( mg \sin \theta \), and the perpendicular is \( mg \cos \theta \).These components are critical in determining how much kinetic energy transforms into potential energy as the jar moves up.
Energy Conservation
Energy conservation is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form to another. This principle is key in solving problems related to motion and forces. In this scenario with the cookie jar on the incline:
  • Initial kinetic energy (KE) is given by \( \frac{1}{2}mv^2 \)
  • Work done against friction reduces the energy
  • Potential energy (PE) is highest at the top of the incline
The equation for energy conservation:\[\frac{1}{2}mv^2 - f_k d = mgh\]By setting up, we account for the work done by friction and the potential energy change. This helps compute how much farther the jar advances before coming to a stop. Overall, energy converts between KE and PE, minus any energy dissipated by friction.
Potential Energy
Potential energy is the energy stored in an object due to its position or height. When dealing with an inclined plane, potential energy comes into play significantly. The jar gains potential energy as it moves up the incline, given by the formula:\[PE = mgh\]where:
  • \( m \) is mass
  • \( g \) is acceleration due to gravity, usually \( 9.81 \, \text{m/s}^2 \)
  • \( h \) is the height the jar reaches
The higher an object goes, the more potential energy it gains. As the jar moves back down, this potential energy converts back into kinetic energy. So as friction decreases, more potential energy translates back to kinetic energy, resulting in a higher speed at the bottom.

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Most popular questions from this chapter

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