Problem 62 - College Physics

Question

A \(1.00-\mathrm{kg}\) block \((M)\) is on a frictionless, \(20^{\circ}\) inclined plane. The block is attached to a spring \((k=25 \mathrm{~N} / \mathrm{m})\) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 40.0 - \(\mathrm{g}\) suspended mass. The suspended mass is given an initial downward speed of \(1.50 \mathrm{~m} / \mathrm{s}\). How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)

Step-by-Step Solution

Answer

The suspended mass drops approximately 0.0464 meters before coming to rest.

Step 1: Understand the System

The system is composed of a block on a frictionless inclined plane, connected to a spring and a hanging mass. The forces involved are gravitational forces, the spring force, and the tension in the string connecting the two masses.

Step 2: Identify the Forces

For the block on the incline:- The force of gravity component parallel to the incline is given by \(M g \sin(\theta)\).- The component normal to the incline is \(M g \cos(\theta)\), which does not affect the motion.- The spring force is given by \(kx\), where \(x\) is the compression/stretch of the spring.For the hanging mass \(m\):- The force of gravity is \(mg\), acting downward.- Tension \(T\) in the string acts upward.

Step 3: Apply Newton's Second Law

For the block on the incline:\[ T + kx = M a + M g \sin(\theta) \]For the hanging mass:\[ mg - T = ma \]

Step 4: Solve for Acceleration and Displacement

Adding the equations from Step 3 together to eliminate \(T\):\[ mg - kx = M a + m a + M g \sin(\theta) \]Using the given values:\[ 0.04g - 25x = 1.04a + 0.34 \]Simplifying into a single motion equation:\[ 0.392 - 25x = 1.04a + 0.34 \]

Step 5: Analyze Mechanical Energy Conservation

Initially, the total energy consists of the kinetic energy of the hanging mass since it has an initial speed:\[ K_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.04 \times 1.5^2 \]When the mass comes to a stop, the energy is stored entirely in the spring as potential energy, balanced by the work done against gravity.\[ U_f = \frac{1}{2} k x^2 \]Equating the initial kinetic energy to the spring potential energy:\[ \frac{1}{2} \times 0.04 \times 1.5^2 = \frac{1}{2} \times 25 \times x^2 + 0.04g x \]

Step 6: Solve for Spring Stretch \(x\)

\[ 0.045 = 12.5x^2 + 0.392x \]Rearrange into a quadratic equation:\[ 12.5x^2 + 0.392x - 0.045 = 0 \]Use the quadratic formula to solve for \(x\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]with \(a = 12.5\), \(b = 0.392\), and \(c = -0.045\).

Step 7: Find the Solution

Calculate the discriminant:\[ b^2 - 4ac = 0.392^2 - 4 \times 12.5 \times (-0.045) \approx 0.1544 + 2.25 = 2.4044 \]Find roots using the positive root since displacement cannot be negative:\[ x = \frac{-0.392 + \sqrt{2.4044}}{2 \times 12.5} \approx \frac{-0.392 + 1.55}{25} \approx 0.0464 \mathrm{~m} \]

Step 8: Conclusion

The suspended mass drops approximately 0.0464 meters before coming to rest, corresponding to the extent to which the spring stretches.

Key Concepts

Mechanical Energy Conservation

The principle of mechanical energy conservation is pivotal in solving this exercise. It tells us that energy in a closed system is neither created nor destroyed; instead, it transforms from one form to another. In this scenario, the hanging mass starts with kinetic energy due to its initial downward speed.
As it descends, this kinetic energy is transferred into potential energy stored in the spring.

• The initial kinetic energy can be expressed as: \( K_i = \frac{1}{2} m v^2 \)
• As the system evolves, the spring stores the energy as potential energy:\( U_f = \frac{1}{2} k x^2 \)

At the point when the hanging mass comes to a stop, all its initial kinetic energy has been transformed into spring potential energy and gravitational potential energy. This concept helps us calculate how far the mass drops by equating the kinetic energy to the energy stored in the spring and gravitational work.

Spring Force

Spring force plays a crucial role in this exercise. Springs exert a force proportional to their compression or extension from the equilibrium position. This force is given by Hooke's Law, formulated as:

• Spring force, \( F_s = kx \)

where \( k \) is the spring constant, a measure of its stiffness, and \( x \) is the distance by which the spring is stretched or compressed. In our scenario, the spring connects to a block sliding on an incline.
As the hanging mass descends, it pulls on the block, stretching the spring. The force from the spring resists this motion.

• This resistance accumulates potential energy in the spring until it equals the energy from the initial motion of the mass.

Understanding how the spring force operates is essential for determining the drop distance of the hanging mass.

Inclined Plane Dynamics

The dynamics of a block on an inclined plane involve analyzing the forces acting on it. Inclined planes are surfaces tilted at a specific angle, which makes the motion analysis different from horizontal surfaces.
For the block on the plane, key forces include:

• Gravity's component parallel to the incline, \( Mg \sin(\theta) \)
• Normal force, \( Mg \cos(\theta) \), acting perpendicular to the surface

Despite the absence of friction in this exercise, these components influence how the block moves along the incline.
The spring force alters the dynamics, as it counters the downward component of gravity, making this a classic physics problem merging incline dynamics and spring mechanics.
Understanding the angle's role helps in balancing these forces and predicting the system's behavior.

Gravitational Forces

Gravitational forces are fundamental to the movement in this problem. For any object on earth, gravity acts downwards, causing acceleration. Here, two masses are in play:

• The block on the inclined plane, where gravity has a parallel and a perpendicular component
• The hanging mass, experiencing direct downward gravitational pull

For the block, the gravitational force’s parallel component, \( Mg \sin(\theta) \), tries to accelerate it downwards. But because of the spring force, its influence is a bit thwarted.
As for the hanging mass, gravity’s action causes it to descend until it loses kinetic energy to the spring.
This interplay of forces dictates how energy moves through the system, providing a real-world glimpse of how objects interact with gravity and other forces.