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Magazine Chapter 10: Problem 23
Block on Incline Collides with Spring In Fig. \(10-37\), a \(12 \mathrm{~kg}\) block is released from rest on a \(30^{\circ}\) frictionless incline. Below the block is a spring that can be compressed \(2.0 \mathrm{~cm}\) by a force of \(270 \mathrm{~N}\). The block momentarily stops when it compresses the spring by \(5.5 \mathrm{~cm} .\) (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches spring?
Short Answer
Expert verified
(a) 0.346 m, (b) 1.84 m/s
Step by step solution
01
Determine the spring constant (k)
Use Hooke's Law to find the spring constant \[ k = \frac{F}{x} \]where \(F = 270 \text{ N}\) and \(x = 2.0 \text{ cm} = 0.02 \text{ m}\). Thus,\[ k = \frac{270}{0.02} = 13500 \text{ N/m} \]
02
Calculate the potential energy stored in the spring (U_s)
When the spring is compressed by \(5.5 \text{ cm}\), the potential energy stored in the spring can be given by \[ U_s = \frac{1}{2} k x^2 \]where \(x = 5.5 \text{ cm} = 0.055 \text{ m}\) and \(k = 13500 \text{ N/m}\).Hence,\[ U_s = \frac{1}{2} \times 13500 \times (0.055)^2 = 20.36 \text{ J} \]
03
Calculate the gravitational potential energy lost by the block (U_g)
The gravitational potential energy lost by the block \(U_g\) is equal to the potential energy stored in the spring \(U_s\). Using \(U_g = mgh\), where \(m = 12 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h\) is the vertical height change.So,\[ 20.36 = 12 \times 9.8 \times h \]Solving for \(h\),\[ h = \frac{20.36}{12 \times 9.8} = 0.173 \text{ m} \]
04
Determine the distance down the incline
Using the relation \(h = d \sin \theta\), where \(d\) is the distance down the incline and \(\theta = 30^{\circ}\), we get\[ d = \frac{h}{\sin 30^{\circ}} = \frac{0.173}{0.5} = 0.346 \text{ m} \]
05
Calculate the speed just before touching the spring
By converting the initial potential energy at the higher point to kinetic energy just before the block touches the spring, we use the principle of conservation of energy:\[ mgh = \frac{1}{2} mv^2 \]Here, \(h\) again will be determined by the inclined distance moved.\[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.173} = 1.84 \text{ m/s} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's laws of motion
Newton's laws of motion are fundamental principles that describe the relationship between the motion of an object and the forces acting on it.
In this problem, we primarily focus on Newton鈥檚 second law, which states: \(F = ma\). This law tells us the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a).
This concept is crucial in understanding how the gravitational force acts on the block along the incline and how this force changes the block鈥檚 motion.
The block鈥檚 acceleration due to gravity can be split into two components:
For the block on the 30掳 incline, we calculate the net force causing the block to move down the slope using the equation: \(F = mg \sin\theta\), where:
\textbf{m} is mass; \(12 \text{ kg}\)
\textbf{g} is acceleration due to gravity; \(9.8 \text{ m/s}^2\)
\textbf{胃} is the angle of the incline; \(30^{\text{掳}}\)
Utilizing the trigonometric relationship, we can analyze the block's motion effectively.
In this problem, we primarily focus on Newton鈥檚 second law, which states: \(F = ma\). This law tells us the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a).
This concept is crucial in understanding how the gravitational force acts on the block along the incline and how this force changes the block鈥檚 motion.
The block鈥檚 acceleration due to gravity can be split into two components:
- \textbf{Parallel to the incline}: This component causes the block to slide down.
- \textbf{Perpendicular to the incline}: This component is balanced by the normal force.
For the block on the 30掳 incline, we calculate the net force causing the block to move down the slope using the equation: \(F = mg \sin\theta\), where:
\textbf{m} is mass; \(12 \text{ kg}\)
\textbf{g} is acceleration due to gravity; \(9.8 \text{ m/s}^2\)
\textbf{胃} is the angle of the incline; \(30^{\text{掳}}\)
Utilizing the trigonometric relationship, we can analyze the block's motion effectively.
conservation of energy
The principle of conservation of energy is pivotal in solving this problem.
This principle states that energy cannot be created or destroyed, only transformed from one form to another.
Here, we have two crucial energy types:
As the block slides down the incline, its gravitational potential energy is converted into two forms:
The energy transformations can be summarized by the equation: \(U_g = \frac{1}{2} k x^2\), where the gravitational potential energy lost equals the elastic potential energy gained by the spring.\
This allows us to calculate how far the block moves down the incline and its speed at different points.
This principle states that energy cannot be created or destroyed, only transformed from one form to another.
Here, we have two crucial energy types:
- \textbf{Gravitational potential energy (U_g)}: Energy stored due to the block鈥檚 height.
- \textbf{Elastic potential energy (U_s)}: Energy stored in the spring when it is compressed by the block.
As the block slides down the incline, its gravitational potential energy is converted into two forms:
- Kinetic energy just before it touches the spring.
- Elastic potential energy when the spring is compressed.
The energy transformations can be summarized by the equation: \(U_g = \frac{1}{2} k x^2\), where the gravitational potential energy lost equals the elastic potential energy gained by the spring.\
This allows us to calculate how far the block moves down the incline and its speed at different points.
inclined plane
An inclined plane is a flat supporting surface tilted at an angle, with one end higher than the other.
In this exercise, the block slides down a 30掳 frictionless incline. The incline makes the problem interesting as it involves analyzing the forces and energy conservation in an angled setup.
The key to solving problems on an inclined plane is to decompose the forces into components parallel and perpendicular to the plane:
By understanding these components, we can move on to applying Newton's laws and energy principles to find distances and velocities.
In this exercise, the block slides down a 30掳 frictionless incline. The incline makes the problem interesting as it involves analyzing the forces and energy conservation in an angled setup.
The key to solving problems on an inclined plane is to decompose the forces into components parallel and perpendicular to the plane:
- \textbf{Parallel component:} This causes the block to slide down and is calculated using \(mg \sin\theta\).
- \textbf{Perpendicular component:} This is balanced by the normal force acting on the block and is calculated using \(mg \cos\theta\).
By understanding these components, we can move on to applying Newton's laws and energy principles to find distances and velocities.
spring constant calculation
The spring constant (k) is a fundamental property of a spring that relates the force needed to compress or extend the spring to the amount it is compressed or extended.
It is calculated using Hooke鈥檚 Law, which states: \(F = kx\), where:
In this problem, we use the given force \(270 \text{ N}\) and compression (\text{2.0} \text{ cm} or \text{0.02} \text{ m}) to calculate the spring constant: \(k = \frac{F}{x} = \frac{270}{0.02} = 13500 \text{ N/m}\).
This spring constant is then used to determine the elastic potential energy stored in the spring when it is further compressed by \text{5.5} \text{ cm}.
It is calculated using Hooke鈥檚 Law, which states: \(F = kx\), where:
- \textbf{F} is the force applied to the spring.
- \textbf{x} is the displacement or compression.
In this problem, we use the given force \(270 \text{ N}\) and compression (\text{2.0} \text{ cm} or \text{0.02} \text{ m}) to calculate the spring constant: \(k = \frac{F}{x} = \frac{270}{0.02} = 13500 \text{ N/m}\).
This spring constant is then used to determine the elastic potential energy stored in the spring when it is further compressed by \text{5.5} \text{ cm}.
potential energy
Potential energy is the stored energy of an object because of its position or configuration.
There are two types relevant to this problem:
As the block slides down the incline, it loses gravitational potential energy, which is converted into elastic potential energy when the spring is compressed.
This conservation of energy principle allows us to solve for different variables such as distance, speed, and spring compression.
There are two types relevant to this problem:
- \textbf{Gravitational potential energy (U_g)}: Given by the formula \(U_g = mgh\), where
- \textbf{Elastic potential energy (U_s)}: Stored in the spring when it is compressed. Calculated using the formula \(U_s = \frac{1}{2} k x^2\).
As the block slides down the incline, it loses gravitational potential energy, which is converted into elastic potential energy when the spring is compressed.
This conservation of energy principle allows us to solve for different variables such as distance, speed, and spring compression.
