Question

The $1.0kg$physics book in FIGURE P$7.41$is connected by a string to a $500g$coffee cup. The book is given a push up the slope and released with a speed of $3.0m/s$. The coefficients of friction are ${\mu }_s=0.50$and ${\mu }_k=0.20$

a. How far does the book slide?

b. At the highest point, does the book stick to the slope, or does it slide back down?

Short answer

(a) The book will slide to a distance of $0.668m$

(b) The book will slide down back.

Step-by-step solution

Given information (part a)

The mass of the book, $m_b=1kg$

The mass of the cup, $m_c=0.5kg$

Initial speed, $u=3m/s$

Coefficient of static friction, ${\mu }_s=0.5$

Coefficient of kinetic friction,${\mu }_k=0.2$

Explanation (part a)

From the free body diagram of the cup, the net forces acting on the cup are Tension (T), weight, and applied force.

$T-m_{c}g=-m_{c}a$

The net forces acting on the book can be shown as

The vertical component can be given as

$$\begin{gathered} F_y=n_b-m_{b}g\cos \left(\theta \right) \\ 0=n_b-m_{b}g\cos \left(\theta \right) \\ {n}_b=m_{b}g\cos \left(\theta \right) \end{gathered}$$

The horizontal component:

localid="1649163010257" $$\begin{gathered} -m_{b}g\sin \left(\theta \right)-{\mu }_k{n}_b-T=m_{b}a \\ a=\frac{m_{b}g(\sin \left(\theta \right)+{\mu }_k\cos \left(\theta \right))+m_{c}g}{m_b+m_c} \\ a=6.73m/s^2 \end{gathered}$$

Equation of kinematics,

localid="1649649521682" $$\begin{gathered} v^2=u^2-2a∆x \\ ∆x=\frac{v^2-u^2}{2a} \\ =\frac{0-{\left(3m/s\right)}^2}{2\left(6.73m/s^2\right)} \\ =0.668m \end{gathered}$$

The book will slide to a distance of $0.668m$

Given information (part b)

The mass of the book, $m_b=1kg$

The mass of the cup, $m_c=0.5kg$

Initial speed, $u=3m/s$

Coefficient of static friction, ${\mu }_s=0.5$

Coefficient of kinetic friction, ${\mu }_k=0.2$

Explanation (part b)

Whether the book sticks to the slope or slides down back, can be known by comparing the maximum amount of static friction to the tension and gravity acting down.

For maximum static friction

$$\begin{gathered} F_s={\mu }_s{n}_b \\ ={\mu }_s{m}_{b}g\cos \left(\theta \right) \\ =0.2\left(1\right)\left(9.8\right)\cos \left(20°\right) \\ =4.91N \end{gathered}$$

For tension

$$\begin{gathered} T=m_{c}g \\ =4.91N \end{gathered}$$

For gravitational pull

localid="1649649548578" $$\begin{gathered} F_g=m_{b}g\sin \left(\theta \right) \\ =\left(1kg\right)\left(9.8m/s^2\right)\sin \left(20°\right) \\ =3.34N \end{gathered}$$

Now, $F_s<\left(T+F_g\right)$

We can see that the total of the forces that causes the downward pull to the book exceed the maximum static friction, hence the book will slide down.