Question

A 10.0-kg block is released from rest at point A in Figure P8.63. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.

Short answer

The coefficient of kinetic friction between the block and the rough surface between points B and C is${\mu }_k=\mathrm{0.328.}$

Step-by-step solution

Step 1:

Friction is adissipative force that acts between two bodies in contact, when one body is in motion relative to the other body. The formula of frictional force is mathematically presented as ${\mathbf{f}}_k\mathbf{=}{\mathbf{\mu }}_k\mathbf{mg}.$:

Here:

$f_k=$Frictional force

$m=$Massof body

$g=$ Gravitational acceleration

Given Data

• Mass of the block$m=10\text{\hspace{0.17em}kg}$
• Force constant of the spring$k=2250\text{\hspace{0.17em}N}/\text{m}$
• The compression in string $x=0.300\text{\hspace{0.17em}m}$

Calculation

Let the point of maximum compression of the spring be D. By applying the law of conservation of energy at various positions given in the diagram,we can further solve the problem.

There is a zerospring potential energy at situation A and a zero gravitational potential energy at situation D, and kinetic energy is zero at points A and D. Putting the energy equation into symbols, we get:

$K_D-K_A-U_A+U_D=-f_k{d}_{BC}$

Here $K$, is the kinetic energy and $U$is the potential energy. Also, the subscripts denote the position at which the quantity is being measured.

Inserting the values of potential and kinetic energy at points $A\text{\hspace{0.17em}and\hspace{0.17em}}D$

$0-0-mg{y}_A+\frac{1}{2}k{x}_s^2=-f_k{d}_{BC}$

Thus, the friction force ismathematically presented as $f_k={\mu }_{k}mg.$

So, $mg{y}_A-\frac{1}{2}k{x}_s^2={\mu }_{k}mgd.$

Solving for the unknown variable ${\mu }_k$ gives:

$\begin{array}{l}{\mu }_k=\frac{y_A}{d}-\frac{k{x}^2}{2mgd}\\ =\frac{3.00\text{m}}{6.00\text{m}}-\frac{(2250\text{N/m}){(0.300\text{m})}^2}{2(10.0\text{kg})(9.8{\text{m/s}}^{\text{2}})(6.00\text{m})}\\ =\mathrm{0.328.}\end{array}$

Therefore, the coefficient of kinetic friction is ${\mu }_k=0.328$