Q6-41E - University Physics with Modern Physics

Question

(a) In Example 6.7 (Section 6.3), it was calculated that with the air track turned off, the glider travels 8.6 cm before it stops instantaneously. How large would the coefficient of static friction ${\mathit{\mu }}_{\mathbf{s}}$have to be to keep the glider from springing back to the left? 

(b) If the coefficient of static friction between the glider and the track is ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.60}$, what is the maximum initial speed${\mathit{v}}_{\mathbf{1}}$ that the glider can be given and still remain at rest after it stops instantaneously? With the air track turned off, the coefficient of kinetic friction is${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.47}$ .

Step-by-Step Solution

Answer

a) The coefficient of static friction to move the glider back to the left is $1.76$.

b) The maximum initial velocity of the glider is $0.67\text{\hspace{0.17em}m/s}$.

 

Step 1: Identification of given data

The given data from example 6.7 can be listed below,

• The coefficient of static friction is,${\mu }_s=0.60$.
• The coefficient of kinetic friction is,${\mu }_k=0.47$.
• The force constant of spring is,$k=20\text{\hspace{0.17em}N/m}$.

·         The mass of the glider is$m=0.100\text{\hspace{0.17em}kg}$,.

Step 2: Concept/Significance of kinetic friction

The frictional force between two bodies that are in touch and moving concerning one another is known as kinetic friction. This motion is opposed by kinetic friction.

Step 3: (a) Determination of the coefficient of static friction μ s have to be to keep the glider from springing back to the left

The static friction in the x-direction is given by,

 $f_s={\mu }_{s}mg$

 

Here, i${\mu }_s$s the coefficient of static friction, m is the mass of the glider, and g is the acceleration due to gravity.

 

The total force on x-direction is given by,

 $f_s-F_{spring}=0$

Substitute all the values in the above to calculate the coefficient of the kinetic friction,

$\begin{array}{c}{\mu }_{s}mg-kd=0\\ {\mu }_s=\frac{kd}{mg}\\ =\frac{\left(20.0\text{\hspace{0.17em}N/m}\right)\left(0.086\text{\hspace{0.17em}m}\right)}{\left(0.100\text{\hspace{0.17em}kg}\right)\left(9.80{\text{\hspace{0.17em}m/s}}^2\right)}\\ =1.76\end{array}$ 

Thus, the coefficient of static friction to move the glider back to the left is $1.76$.

 

Step 4: (b) Determination of the maximum initial speed v 1 that the glider can be given and still remain at rest after it stops instantaneously

The distance traveled by the glider is given by,

 $\begin{array}{l}{\mu }_{s}mg=kd\\ d=\frac{{\mu }_{s}mg}{k}\end{array}$

 

Here,${\mu }_s$ is the coefficient of static friction, m is the mass of the glider and g is the acceleration due to gravity, k is the force constant.

 

Substitute all the values in the above,

 $\begin{array}{c}d=\frac{\left(0.60\right)\left(0.100\right)\left(9.80{\text{\hspace{0.17em}m/s}}^2\right)}{20\text{\hspace{0.17em}N/m}}\\ =0.0294\text{\hspace{0.17em}m}\end{array}$

 

The total work done by the glider is given by,

 $\begin{array}{c}{W}_{tot}=K_f-K_i\\ =\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2\end{array}$

 

The total work done by the system is given by,

 $\begin{array}{c}{W}_{tot}=W_{spring}+W_{fric}\\ =-\frac{1}{2}k{d}^2-{\mu }_{k}mgd\end{array}$

 

Substitute all the values in the above,

 $\begin{array}{c}{W}_{tot}=-\frac{1}{2}\left(20\text{\hspace{0.17em}N/m}\right){\left(0.0294\text{\hspace{0.17em}m}\right)}^2-0.47\left(0.100\text{\hspace{0.17em}kg}\right)\left(9.80{\text{\hspace{0.17em}m/s}}^2\right)\left(0.0294\text{\hspace{0.17em}m}\right)\\ =-0.022\text{\hspace{0.17em}J}\end{array}$

Substitute the value of work done to calculate the velocity of the glider,

 $\begin{array}{c}-0.22\text{\hspace{0.17em}J}=0-\frac{1}{2}\left(0.100\text{\hspace{0.17em}kg}\right)v_1^2\\ v_1=\sqrt{\frac{2\times 0.22\text{\hspace{0.17em}J}}{0.100\text{\hspace{0.17em}kg}}}\\ =0.67\text{\hspace{0.17em}m/s}\end{array}$

 

Thus, the maximum initial velocity of the glider is$0.67\text{\hspace{0.17em}m/s}$.