a) Find D in terms of m, k, L, g, and 0. b) Describe in words what the motion will look like over time.
a) Find D in terms of m, k, L, g, and 0. b) Describe in words what the motion will look like over time.
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Question
![**Inclined Plane with Spring**
**Problem Description:**
A block of mass \( m \) is at the top of an inclined plane of length \( L \) and angle \( \theta \). At the bottom of the plane, there is a relaxed spring with spring constant \( k \) (See the figure). The block is released from rest and slides without friction down the plane. At the bottom of the plane, the block strikes the spring and momentarily comes to rest by compressing the spring by some amount \( D \).
**Tasks:**
(a) Find \( D \) in terms of \( m \), \( k \), \( L \), \( g \), and \( \theta \).
(b) Describe in words what the motion will look like over time.
**Diagram Explanation:**
The diagram is divided into two parts. In the top part, a block is shown at the top of an inclined plane, labeled with angle \( \theta \), and an arrow indicates the path it will take down the plane. In the lower part, the block is shown compressing a spring at the bottom of the incline. The inclination and compression of the spring are illustrated, and the spring compression is labeled as "temporarily at rest."
**Analysis:**
The setup involves energy conservation principles. Initially, the block has potential energy due to its height on the incline, which gets converted into kinetic energy as it slides down. When the block compresses the spring, kinetic energy is converted into spring potential energy.
**Mathematical Solution:**
To solve part (a), you need to apply energy conservation as follows:
1. The gravitational potential energy at the top is given by:
\( PE = mgh \),
where \( h = L \sin \theta \).
2. When the block compresses the spring maximally, all the gravitational potential energy is transferred to the spring:
\[
\frac{1}{2} k D^2 = mgL \sin \theta
\]
3. Solve for \( D \):
\[
D = \sqrt{\frac{2mgL \sin \theta}{k}}
\]
**Motion Description (b):**
The block starts from rest and accelerates down the incline due to gravity. As it descends, it speeds up and reaches the spring with maximum kinetic energy. Upon compressing the spring, the block's kinetic energy is transferred to the spring,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8db46b6b-83ec-4d74-9974-6a06215af567%2Fe5685fcc-728b-4e7a-9bf6-f06d063a03c2%2Fcg868sk_processed.png&w=3840&q=75)
Transcribed Image Text:**Inclined Plane with Spring**
**Problem Description:**
A block of mass \( m \) is at the top of an inclined plane of length \( L \) and angle \( \theta \). At the bottom of the plane, there is a relaxed spring with spring constant \( k \) (See the figure). The block is released from rest and slides without friction down the plane. At the bottom of the plane, the block strikes the spring and momentarily comes to rest by compressing the spring by some amount \( D \).
**Tasks:**
(a) Find \( D \) in terms of \( m \), \( k \), \( L \), \( g \), and \( \theta \).
(b) Describe in words what the motion will look like over time.
**Diagram Explanation:**
The diagram is divided into two parts. In the top part, a block is shown at the top of an inclined plane, labeled with angle \( \theta \), and an arrow indicates the path it will take down the plane. In the lower part, the block is shown compressing a spring at the bottom of the incline. The inclination and compression of the spring are illustrated, and the spring compression is labeled as "temporarily at rest."
**Analysis:**
The setup involves energy conservation principles. Initially, the block has potential energy due to its height on the incline, which gets converted into kinetic energy as it slides down. When the block compresses the spring, kinetic energy is converted into spring potential energy.
**Mathematical Solution:**
To solve part (a), you need to apply energy conservation as follows:
1. The gravitational potential energy at the top is given by:
\( PE = mgh \),
where \( h = L \sin \theta \).
2. When the block compresses the spring maximally, all the gravitational potential energy is transferred to the spring:
\[
\frac{1}{2} k D^2 = mgL \sin \theta
\]
3. Solve for \( D \):
\[
D = \sqrt{\frac{2mgL \sin \theta}{k}}
\]
**Motion Description (b):**
The block starts from rest and accelerates down the incline due to gravity. As it descends, it speeds up and reaches the spring with maximum kinetic energy. Upon compressing the spring, the block's kinetic energy is transferred to the spring,
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