A 7.50-kg box initially at the horizontal (see figure). At the bottom of a) Draw a free body diagram, detailing all of the forces acting on the box. 0 ansion for the
A 7.50-kg box initially at the horizontal (see figure). At the bottom of a) Draw a free body diagram, detailing all of the forces acting on the box. 0 ansion for the
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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![### Introduction to Kinetic Friction on an Incline
1. A 7.50-kg box initially at rest slides down a 15.0 m long incline that makes an angle of 22.0° with the horizontal (see figure). At the bottom of the incline, the speed of the box is 3.00 m/s.
#### a) Draw a free body diagram, detailing all of the forces acting on the box.
To begin our analysis, construct a free body diagram that shows all the forces acting on the box:
- **Gravity (Weight) Force (mg)**: Acts vertically downward.
- **Normal Force (N)**: Acts perpendicular to the surface of the incline.
- **Kinetic Friction Force (f_k)**: Acts parallel to the surface of the incline and opposite to the direction of motion.
The diagram on the right shows a box on an inclined plane making an angle \(\theta\) with the horizontal.
#### b) Using the work done by non-conservative forces relationship, find the algebraic expression for the coefficient of kinetic friction between the box and the incline.
To find the coefficient of kinetic friction (\(\mu_k\)), we can use the principle of work and energy. The work done by non-conservative forces (in this case, kinetic friction) affects the mechanical energy of the system.
\[ W_{nc} = \Delta K + \Delta U_g \]
Where:
- \( W_{nc} \) is the work done by non-conservative forces.
- \( \Delta K \) is the change in kinetic energy.
- \( \Delta U_g \) is the change in gravitational potential energy.
Initial kinetic energy (\( K_i \)) is zero since the box starts from rest:
\[ K_i = 0 \]
Final kinetic energy (\( K_f \)) is:
\[ K_f = \frac{1}{2}mv^2 \]
Change in kinetic energy:
\[ \Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2 \]
Change in gravitational potential energy:
\[ \Delta U_g = mg(h_f - h_i) = -mgh \]
The height \( h \) can be found using the sine function:
\[ h = L \sin(\theta) \]
So:
\[ \Delta U_g = -mgL\sin(\theta) \]
By combining the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c6fb9be-cb3b-43fd-a6cc-90b4ac965f54%2Fe5897d13-878d-4afd-9549-a3c3af02a7f7%2Fivk7r0o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Introduction to Kinetic Friction on an Incline
1. A 7.50-kg box initially at rest slides down a 15.0 m long incline that makes an angle of 22.0° with the horizontal (see figure). At the bottom of the incline, the speed of the box is 3.00 m/s.
#### a) Draw a free body diagram, detailing all of the forces acting on the box.
To begin our analysis, construct a free body diagram that shows all the forces acting on the box:
- **Gravity (Weight) Force (mg)**: Acts vertically downward.
- **Normal Force (N)**: Acts perpendicular to the surface of the incline.
- **Kinetic Friction Force (f_k)**: Acts parallel to the surface of the incline and opposite to the direction of motion.
The diagram on the right shows a box on an inclined plane making an angle \(\theta\) with the horizontal.
#### b) Using the work done by non-conservative forces relationship, find the algebraic expression for the coefficient of kinetic friction between the box and the incline.
To find the coefficient of kinetic friction (\(\mu_k\)), we can use the principle of work and energy. The work done by non-conservative forces (in this case, kinetic friction) affects the mechanical energy of the system.
\[ W_{nc} = \Delta K + \Delta U_g \]
Where:
- \( W_{nc} \) is the work done by non-conservative forces.
- \( \Delta K \) is the change in kinetic energy.
- \( \Delta U_g \) is the change in gravitational potential energy.
Initial kinetic energy (\( K_i \)) is zero since the box starts from rest:
\[ K_i = 0 \]
Final kinetic energy (\( K_f \)) is:
\[ K_f = \frac{1}{2}mv^2 \]
Change in kinetic energy:
\[ \Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2 \]
Change in gravitational potential energy:
\[ \Delta U_g = mg(h_f - h_i) = -mgh \]
The height \( h \) can be found using the sine function:
\[ h = L \sin(\theta) \]
So:
\[ \Delta U_g = -mgL\sin(\theta) \]
By combining the
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