An 0.8kg block is held in place against a spring with a spring constant of 3900 J/m², compressing it by 1.7 cm. The spring is located on a platform elevated 11cm above the ground. The block is released, slides across the slick platform and down a ramp and across the floor. The floor has a rough patch 30cm long. If the block has a speed of 0.8m/s after going across the rough patch, how much energy was dissipated due to friction?
An 0.8kg block is held in place against a spring with a spring constant of 3900 J/m², compressing it by 1.7 cm. The spring is located on a platform elevated 11cm above the ground. The block is released, slides across the slick platform and down a ramp and across the floor. The floor has a rough patch 30cm long. If the block has a speed of 0.8m/s after going across the rough patch, how much energy was dissipated due to friction?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please use equations given on equation sheet
![Below is a transcription and explanation of key physics equations displayed in the image. These are fundamental concepts typically taught in physics courses:
1. **Momentum Equation:**
\[
\mathbf{p} = m\mathbf{v}
\]
- This equation defines momentum (\(\mathbf{p}\)) as the product of mass (\(m\)) and velocity (\(\mathbf{v}\)).
2. **Conservation of Linear Momentum:**
\[
\Sigma \mathbf{p}_{n,i} = \Sigma \mathbf{p}_{n,f}
\]
- This equation represents the conservation of momentum. The initial total momentum (\(\Sigma \mathbf{p}_{n,i}\)) equals the final total momentum (\(\Sigma \mathbf{p}_{n,f}\)) in a closed system.
3. **Impulse:**
\[
\mathbf{J} = \Delta \mathbf{p}
\]
- Impulse (\(\mathbf{J}\)) is equal to the change in momentum (\(\Delta \mathbf{p}\)).
4. **Coefficient of Restitution:**
\[
e = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}
\]
- This formula calculates the coefficient of restitution (\(e\)), which is the ratio of relative velocities after and before an impact, indicating how elastic a collision is.
5. **Kinetic Energy:**
\[
K = \frac{1}{2}mv^2
\]
- This formula defines kinetic energy (\(K\)) as half the product of mass (\(m\)) and velocity squared (\(v^2\)).
6. **Gravitational Potential Energy:**
\[
U^G = mgh
\]
- Gravitational potential energy (\(U^G\)) is determined by mass (\(m\)), gravitational acceleration (\(g\)), and height (\(h\)).
7. **Elastic Potential Energy:**
\[
U^S = \frac{1}{2}k\Delta x^2
\]
- Elastic potential energy (\(U^S\)) is given by half the product of the spring constant (\(k\)) and the square of the displacement (\(\Delta x^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc17bbd9d-3402-481c-8852-0ad413726a14%2F10f141d2-afa5-4649-a8f5-3eb8a1d22e12%2Fsi6mfh5_processed.png&w=3840&q=75)
Transcribed Image Text:Below is a transcription and explanation of key physics equations displayed in the image. These are fundamental concepts typically taught in physics courses:
1. **Momentum Equation:**
\[
\mathbf{p} = m\mathbf{v}
\]
- This equation defines momentum (\(\mathbf{p}\)) as the product of mass (\(m\)) and velocity (\(\mathbf{v}\)).
2. **Conservation of Linear Momentum:**
\[
\Sigma \mathbf{p}_{n,i} = \Sigma \mathbf{p}_{n,f}
\]
- This equation represents the conservation of momentum. The initial total momentum (\(\Sigma \mathbf{p}_{n,i}\)) equals the final total momentum (\(\Sigma \mathbf{p}_{n,f}\)) in a closed system.
3. **Impulse:**
\[
\mathbf{J} = \Delta \mathbf{p}
\]
- Impulse (\(\mathbf{J}\)) is equal to the change in momentum (\(\Delta \mathbf{p}\)).
4. **Coefficient of Restitution:**
\[
e = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}
\]
- This formula calculates the coefficient of restitution (\(e\)), which is the ratio of relative velocities after and before an impact, indicating how elastic a collision is.
5. **Kinetic Energy:**
\[
K = \frac{1}{2}mv^2
\]
- This formula defines kinetic energy (\(K\)) as half the product of mass (\(m\)) and velocity squared (\(v^2\)).
6. **Gravitational Potential Energy:**
\[
U^G = mgh
\]
- Gravitational potential energy (\(U^G\)) is determined by mass (\(m\)), gravitational acceleration (\(g\)), and height (\(h\)).
7. **Elastic Potential Energy:**
\[
U^S = \frac{1}{2}k\Delta x^2
\]
- Elastic potential energy (\(U^S\)) is given by half the product of the spring constant (\(k\)) and the square of the displacement (\(\Delta x^2
Transcribed Image Text:**Physics Problem: Energy Dissipation Due to Friction**
A \(0.8 \ \text{kg}\) block is held in place against a spring with a spring constant of \(3900 \ \text{J/m}^2\), compressing it by \(1.7 \ \text{cm}\). The spring is located on a platform elevated \(11 \ \text{cm}\) above the ground.
When released, the block slides across the slick platform, travels down a ramp, and moves across the floor. The floor has a rough patch \(30 \ \text{cm}\) long. After passing this rough patch, the block has a speed of \(0.8 \ \text{m/s}\).
The question is: How much energy was dissipated due to friction while the block moved across the rough patch?
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