Q 36E - University Physics with Modern Physics

Question

Question: A 25.0kg box of textbooks rests on a loading ramp that makes an angle $\mathit{\alpha }$ with the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As $\mathit{\alpha }$ is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0m along the loading ramp?

Step-by-Step Solution

Answer

(a) Minimum angle at which box starts to slip is, $19.29°$.

(b) The acceleration of the box is $0.931m/s^2$.

(c) The velocity of the box after slide is 3.05m/s.

Step 1: Identification of given data:

The given data can be listed below as,

• The mass of box of textbook is 25.0kg.
• The coefficient of kinetic friction is ${\mu }_k=0.25$.
• The coefficient of static friction is ${\mu }_s=0.35$.
• Distance of box when slide along the ramp is s = 5.0m.

Step 2: Significance of kinetic friction and static friction:

The kinetic friction is the force of friction that occurs when the object is accelerating and the static friction is the force of friction that occurs when the object is at rest.

Step 3: (a) Identification of minimum angle at which box starts to slip:

In the free body diagram given above, the all forces act on the box can be expressed as,

$m g+F_{fr}+N=0$

Here, m is the mass of the box, g is the acceleration due to gravity and $F_{fr}$ is the frictional force and N is the normal force.

The projected forces of x and y axis are expressed as,

$$\begin{gathered} m g \sin \alpha -F_{fr}=0 \\ N-m g \cos \alpha =0 \\ N=m g \cos \alpha \end{gathered}$$

From the above free body diagram the frictional force act on the box is expressed as,

$\begin{array}{rcl}{F}_{fr}& =& {\mu }_s N\\ & =& {\mu }_s m g \cos \alpha \end{array}$

Here, ${\mu }_s$ is the coefficient of static friction.

 

Substitute the value of $F_{fr}$ and 0.35 for ${\mu }_s$ in the above equation.

 $$\begin{gathered} m g \sin \alpha -{\mu }_s m g \cos \alpha \\ \sin \alpha ={\mu }_s \cos \alpha \\ \tan \alpha =0.35 \\ \alpha ={\text{tan}}^{-1} (0.35) \end{gathered}$$

 $\alpha =19.29°$

 

Hence, the required angle is $19.29°$.

Step 4: (b) Determination of acceleration:

The projection of the forces on x axis can be expressed as,

 $$\begin{gathered} mg \sin \alpha -{\mu }_k mg \cos \alpha =ma \\ mg\left(\sin \alpha -{\mu }_k \cos \alpha \right)=ma \\ a=g\left(\sin \alpha -{\mu }_k \cos \alpha \right) \end{gathered}$$

Here, a is the acceleration, ${\mu }_k$ is the coefficient of kinetic friction.

Substitute $9.8m/s^2$ for g, $19.29°$ for $\alpha$ and 0.25 for ${\mu }_k$ in the above equation.

$\begin{array}{rcl}a& =& \left(9.8m/s^2\right)(\sin 19.29°-0.25\times \cos 19.29°)\\ & =& 0.931m/s^2\end{array}$

Hence, the required acceleration is $0.931m/s^2$.

Step 5: (c) Determination of velocity of box at 5.0m:

The kinematic equation of the velocity is expressed as,

$$\begin{gathered} v^2=v_0^2+2 as \\ v=\sqrt{v_0^2+2 as} \end{gathered}$$ 

Here, v is the velocity of box after slide, $v_0$ is the initial velocity and a is the acceleration and s is the sliding distance.

 

Substitute 0m/s for $v_0$, $0.931m/s^2$ for a, and 5.0m for s in the above equation.

 $\begin{array}{rcl}v& =& \sqrt{0+2 (0.931m/s^2)(5.0m)}\\ & =& 3.05m/s\end{array}$

Hence, the required velocity is 3.05m/s.