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A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\), which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (\(Hint\): This does \(not\) occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

Short Answer

Expert verified
The tray rises 0.014 above point A before the ball leaves. It takes about 0.067 seconds for the ball to leave the tray. The ball moves at 0.677 m/s as it leaves the tray.

Step by step solution

01

Understand the System

The system consists of a tray attached to a spring with a known trigger force constant, a metal ball in the tray, and an initial setup where the tray is displaced 15 cm below the equilibrium point. We need to find when and how the ball separates from the tray as it oscillates due to the spring force.
02

Analyze Forces at Separation

The ball will leave the tray when the upward force from the spring exceeds the gravitational force on the ball. Calculate the gravitational force on the ball: \( F_g = m_b g = 0.275 \times 9.8 = 2.695 \text{ N} \), where \(m_b\) is the mass of the ball.
03

Equilibrium Force Calculation

At equilibrium, the spring force equals the combined weight of the tray and the ball. Calculate the equilibrium position using Hooke's Law for the initial displacement: \( kx = (m_t + m_b)g \) where \(x\) is the compression from equilibrium. Thus, \( x = \frac{(1.50 + 0.275) \times 9.8}{185} \).
04

Effective Spring Force for Separation

Find the spring force needed to lift the ball. The spring force that causes separation equals the gravitational force on the ball: \(F_s = k(x + h) = m_b g\). Solve for \(h\) where separation occurs.
05

Solve for Separation Height

To calculate the height \(h\) above initial point, set \( k(x + h) = 2.695 \). Solve for \(h\): \(h = \frac{2.695}{185} - x\).
06

Time Calculation for Ball Release

Determine the time when the ball leaves the tray by finding the time for the system to travel from initial point to separation point using the formula of harmonic motion: \( T = 2\pi \sqrt{\frac{m_t + m_b}{k}} \) to find period, then derive the quarter period for release timing.
07

Velocity Calculation at Separation

To find the velocity of the ball at separation, use energy conservation: \( \frac{1}{2} m v^2 = \frac{1}{2} k (x + h)^2 \), extract "v" for ball’s velocity just as it leaves.

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Key Concepts

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

harmonic motion
Harmonic motion is a type of repetitive oscillation that occurs when an object moves back and forth around a fixed point, called the equilibrium point. This kind of motion is common in systems like springs, pendulums, and certain types of waves.

In our exercise, the system consists of a tray and a metal ball, which are both attached to a vertical spring. When the spring is stretched or compressed, it exerts a force to return to equilibrium. This creates an oscillatory movement, leading to harmonic motion.

Key characteristics of harmonic motion include:
- The motion is symmetric about the equilibrium point.
- It is periodic, meaning it repeats at regular intervals.
- The restoring force is directly proportional to the displacement from equilibrium, which is described by Hooke's Law.

Understanding these principles is crucial, as they influence the timing of events, like when the ball leaves the tray.
spring force
Spring force is central to the tray-and-ball system. This force is exerted by the spring and works to bring the spring back to its natural, uncompressed state.

According to Hooke's Law, the spring force is calculated as:
\[F_s = k x\]
where \( F_s \) is the spring force, \( k \) is the spring constant (185 N/m in this case), and \( x \) is the displacement from the spring's equilibrium position.

In the exercise, you're looking at how this spring force interacts with other forces such as gravity. When the ball leaves the tray, the spring force working upwards exceeds the downward gravitational force acting on the ball, calculated as:
\[F_g = m_b g = 0.275 imes 9.8 = 2.695 ext{ N}\]
Therefore, understanding how spring force works allows us to determine at which point the ball has enough upward force to lift off from the tray.
mechanical oscillations
Mechanical oscillations refer to the back-and-forth movement that occurs in systems like our tray and metal ball attached to a spring. These oscillations are a form of harmonic motion, characterized by the regular shifting of the system between potential and kinetic energy.

Important points about mechanical oscillations include:
- The period of oscillation, which is the time it takes for one complete cycle of motion. It's influenced by the mass and the stiffness (spring constant) of the system, calculated as:
\[T = 2\pi \sqrt{\frac{m_t + m_b}{k}}\]
where \( m_t + m_b \) is the combined mass of the tray and ball, and \( k \) is the spring constant.
- The amplitude, which is the maximum extent of oscillation from the equilibrium position.

Mechanical oscillations help us determine much about the system's behavior, such as the time it will take for the ball to leave the tray, as it's a quarter of the oscillation period.
energy conservation
Energy conservation is a fundamental concept that plays a pivotal role in understanding oscillatory systems. It states that energy in an isolated system remains constant—it's neither created nor destroyed, only transformed between different types.

In our spring-tray-ball system, energy changes forms between kinetic and potential energy as the spring compresses and expands. Initially, when the tray is pressed down, the system has maximum potential energy. As it is released and moves upwards, the potential energy converts into kinetic energy until reaching another maximum at equilibrium.

To find the velocity of the ball when it leaves the tray, we apply energy conservation as follows:
\[\frac{1}{2} m v^2 = \frac{1}{2} k (x + h)^2\]
where \( m \) is the mass of the ball, \( v \) is its velocity, \( k \) represents the spring constant, and \( x + h \) is the displacement at separation.

This formula helps us calculate energy distribution and, consequently, the speed and point at which the ball departs from the tray.

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Most popular questions from this chapter

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 N \(\cdot\) m/rad. You twist the part a small amount about this axis and let it go, timing 165 oscillations in 265 s. What is its moment of inertia?

A 5.00-kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s. (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 m above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

The point of the needle of a sewing machine moves in SHM along the \(x\)-axis with a frequency of 2.5 Hz. At \(t =\) 0 its position and velocity components are \(+\)1.1 cm and \(-\)15 cm/s, respectively. (a) Find the acceleration component of the needle at \(t =\) 0. (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

At the end of a ride at a winter-theme amusement park, a sleigh with mass 250 kg (including two passengers) slides without friction along a horizontal, snow-covered surface. The sleigh hits one end of a light horizontal spring that obeys Hooke's law and has its other end attached to a wall. The sleigh latches onto the end of the spring and subsequently moves back and forth in SHM on the end of the spring until a braking mechanism is engaged, which brings the sleigh to rest. The frequency of the SHM is 0.225 Hz, and the amplitude is 0.950 m. (a) What was the speed of the sleigh just before it hit the end of the spring? (b) What is the maximum magnitude of the sleigh's acceleration during its SHM?

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