/*! 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 60 Cookie Jar A cookie jar is movin... [FREE SOLUTION] | 91Ó°ÊÓ

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Cookie Jar 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), it 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
Jar travels additional distance up incline before zero speed initially then back to base w/ initial given speed intact+ decrease friction increases distance.

Step by step solution

01

Identify Forces Acting on the Jar

Consider the forces acting on the jar: gravitational force (\(mg\), where \(m\) is the mass of the jar and \(g\) is the acceleration due to gravity), normal force (\(N\)), and kinetic friction force (\(f_k = \) \(\frac{\text{force of friction}}{normal force})\) which equals \( \frac{\text{gravity component parallel to the incline}}{\text{gravity component perpendicular to incline}}\)
02

Calculate the Components of Gravitational Force

The gravitational force can be split into two components: one parallel to the incline (\(m \times g \times \text{sin} \theta\)) and one perpendicular to the incline (\(m \times g \times \text{cos} \theta\)). For the given angle \(\theta = 40^\text{o}\) and gravitational acceleration \(g = 9.8 \text{m}/\text{s}^2\)
03

Determine Normal Force and Kinetic Friction

The normal force \(N = m \times g \times \text{cos} \theta\). The kinetic friction force is \(f_k = \text{coefficient of kinetic friction} \times normal force \) = \(0.15 \times m \times g \times \text{cos} \theta\)
04

Calculate the Net Force Acting on the Jar

Parallel force \(F_{\text{parallel}}\) is the difference between the gravitational force component down the incline and the kinetic friction force. Therefore, \(F_{\text{parallel}} = m \times g \times \text{sin} \theta - f_k\). Substitute \(f_k\) as determined in Step 3.
05

Apply the Work-Energy Principle Upslope

Utilize the Work-Energy Theorem to determine how much farther the jar will travel up the incline. Initial kinetic energy is \(\frac{1}{2} m (1.4)^2\), and work done by the net force is \( W = F_{\text{parallel}} d \) to find additional distance jar will travel. Initial distance is 0.55m.
06

Calculate Distance the Jar Moves Up

Solve \(\frac{1}{2} m (1.4)^2 = (m \times g \times \text{sin} \theta - 0.15 \times m \times g \times \text{cos} \theta) \times d\) for \(d\) with \(\theta = 40^\text{o}\)
07

Determine Final Speed When Jar Slides Back

The jar travels back to the base. Apply energy conservation principles: potential energy gained while moving back downslope to kinetic energy. Hence final velocity \(v_{bottom}\) determined.
08

Change Coefficient of Friction for Analysis

Repeat the analysis with a lower coefficient of kinetic friction to compare jar behavior. The distance traveled uphill would increase when friction coefficient decreases as less energy converted against friction.

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Key Concepts

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

Incline Plane Problems
Incline plane problems are a staple in physics education, often used to illustrate fundamental concepts of forces and motion. When dealing with these problems, you need to understand how the gravitational force splits into components along and perpendicular to the plane's surface.

For instance, in our exercise, we deal with a cookie jar moving up a 40-degree incline. The gravitational force acting on this jar can be divided:

  • The parallel component: This is calculated using the formula for sine: \(m \times g \times \text{sin} \theta \).
  • The perpendicular component: This uses the formula for cosine: \(m \times g \times \text{cos} \theta\).

Knowing these components helps in calculating other forces, like the normal force and friction.
Work-Energy Theorem
A crucial tool in solving incline plane problems is the Work-Energy Theorem. This theorem connects the concepts of work done by forces and the change in kinetic energy of the object.

In simpler terms, the Work-Energy Theorem states: \[W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \]
The net work (\text{W_{\text{net}}}}) done on an object is equal to its change in kinetic energy. Applying this to our exercise, we start with the initial kinetic energy given by \(\frac{1}{2} m (1.4)^2\). The net force (which we calculated using gravitational forces and friction) does work to slow the jar down until it eventually stops.

After determining the work done by the parallel component of gravitational force and friction, we could solve for the additional distance the jar will travel before stopping completely.
Kinetic Energy and Friction
In inclined plane problems, particularly those involving kinetic energy and friction, understanding the role of friction is pivotal.

Kinetic energy is given by \(\frac{1}{2}mv^2\), and as the jar slides up, it loses kinetic energy due to the work done against friction and the slope's gravitational force. Here, friction force \(f_k\) is calculated as:
\(f_k = \text{coefficient of kinetic friction} \times \text{normal force} = 0.15 \times m \times g \times \text{cos} \theta \)
The force of friction opposes the motion, thus removing energy from the system, evidenced by slowing down the jar as it moves up.

When analyzing how the jar slides back down, understanding energy conservation is key. The jar regains kinetic energy lost to friction and potential energy as it moves back to the bottom, allowing us to calculate its speed at the bottom.

Altering the coefficient of kinetic friction impacts the results significantly, with lower friction leading to the jar traveling further up due to less energy loss to friction.

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Most popular questions from this chapter

To Make a Pendulum To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of \(a\) string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

Ski Jumper A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(24 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of \(22 \mathrm{~m} / \mathrm{s}\), landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

Shows a ball with mass \(m\) attached to the end of a thin rod with length \(L\) and negligible mass. The other end of the rod is pivoted so that the ball can move in a vertical circle. The rod is held in the horizontal position as shown and then given enough of a downward push to cause the ball to swing down and around and just reach the vertically upward position, with zero speed there. How much work is done on the ball by the gravitational force from the initial point to (a) the lowest point, (b) the highest point, and (c) the point on the right at which the ball is level with the initial point? If the gravitational potential energy of the ball-Earth system is taken to be zero at the initial point, what is its value when the ball reaches (d) the lowest point, (e) the highest point, and (f) the point on the right that is level with the initial point? (g) Suppose the rod were pushed harder so that the ball passed through the highest point with a nonzero speed. Would the change in the gravitational potential energy from the lowest point to the highest point then be greater, less, or the same?

Billiard Ball A billiard ball moving at a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) strikes an identical stationary ball a glancing blow. After the collision, one ball is found to be moving at a speed of \(1.1 \mathrm{~m} / \mathrm{s}\) in a direction making a \(60^{\circ}\) angle with the original line of motion. (a) Find the velocity of the other ball. (b) Can the collision be inelastic, given these data?

Small Ball Above Larger A small ball of mass \(m\) is aligned above a larger ball of mass \(M\) (with a slight separation, and the two are dropped simultaneously from \(h .\) (Assume the radius of each ball is negligible compared to \(h .)(\mathrm{a})\) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what ratio \(m / M\) results in the larger ball stopping upon its collision with the small ball? (The answer is approximately the mass ratio of a baseball to a basketball.) (b) What height does the small ball then reach?

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