a. Write an expression to describe the scalar component of your pushing force in the x-direction as a function of time. Remember, it has a value of F, at t = 0 and linearly decreases until it reaches 0 at t = 10s. (This step has nothing to do with forces or physics, it is purely mathematical.) b. In which direction does the acceleration of the book point? In which direction does the parallel component of the contact force (a.k.a. the friction force) point? c. Draw a FBD for the book.

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### Problem Set

a. **Expression for Scalar Component of Force in x-direction:**
   - Write an expression to describe the scalar component of your pushing force in the x-direction as a function of time. It starts with a value of \( F_0 \) at \( t = 0 \) and decreases linearly, reaching 0 at \( t = 10s \).

b. **Direction of Acceleration and Friction Force:**
   - Determine the direction of the book's acceleration.
   - Identify the direction of the parallel component of the contact force (i.e., the friction force).

c. **Free Body Diagram (FBD):**
   - Draw a Free Body Diagram (FBD) for the book.

d. **Newton’s 2nd Law Equation:**
   - Write a Newton’s 2nd Law equation for the book, using \( \vec{a} \) as a variable. Note that \( \vec{a} \) varies between 0-5 seconds and 5-10 seconds.
Transcribed Image Text:### Problem Set a. **Expression for Scalar Component of Force in x-direction:** - Write an expression to describe the scalar component of your pushing force in the x-direction as a function of time. It starts with a value of \( F_0 \) at \( t = 0 \) and decreases linearly, reaching 0 at \( t = 10s \). b. **Direction of Acceleration and Friction Force:** - Determine the direction of the book's acceleration. - Identify the direction of the parallel component of the contact force (i.e., the friction force). c. **Free Body Diagram (FBD):** - Draw a Free Body Diagram (FBD) for the book. d. **Newton’s 2nd Law Equation:** - Write a Newton’s 2nd Law equation for the book, using \( \vec{a} \) as a variable. Note that \( \vec{a} \) varies between 0-5 seconds and 5-10 seconds.
**Description:**

You are pushing a book with a force \( \vec{F}_{yb} \) to the right across a level table at a constant velocity under the influence of kinetic friction. At \( t = 0 \, s \), you reduce the magnitude of your force from \( F_o \) to 0, at a constant rate over the course of 10 seconds. 

For the first 5 seconds of this time, the book is still sliding; however, during the last 5 seconds, the book is at rest with respect to the table. 

Use a coordinate system such that the \( \hat{x} \) points to the right. 

**Explanation:**

In this scenario, initially, the applied force is equal to the force of kinetic friction, which allows the book to move at a constant velocity. Over the next 10 seconds, as you gradually decrease the applied force to zero, the opposing friction force reduces the book's velocity to zero in the first 5 seconds. The book remains stationary for the remaining 5 seconds of the interval when no force is applied.
Transcribed Image Text:**Description:** You are pushing a book with a force \( \vec{F}_{yb} \) to the right across a level table at a constant velocity under the influence of kinetic friction. At \( t = 0 \, s \), you reduce the magnitude of your force from \( F_o \) to 0, at a constant rate over the course of 10 seconds. For the first 5 seconds of this time, the book is still sliding; however, during the last 5 seconds, the book is at rest with respect to the table. Use a coordinate system such that the \( \hat{x} \) points to the right. **Explanation:** In this scenario, initially, the applied force is equal to the force of kinetic friction, which allows the book to move at a constant velocity. Over the next 10 seconds, as you gradually decrease the applied force to zero, the opposing friction force reduces the book's velocity to zero in the first 5 seconds. The book remains stationary for the remaining 5 seconds of the interval when no force is applied.
Expert Solution
Step 1

Given,

force is the act in the right direction that is on the x-axis. Where,

Ft=0=FoFt=10=0

And there is a linear relationship between force and time. Then force will be,

F=Fo1010-tx^F=Fo1-t10x^

a)

So the component of Force in x-direction will be,

Fx=Fo1-t10

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