Question

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (Fig. P8.65). the force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point A, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The block’s speed at the bottom of the track is ${\mathbf{v}}_{\mathbf{A}}\mathbf{=}\mathbf{12}\mathbf{.0}\mathbf{}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, and the block experiences an average friction force of 7.00 N while sliding up the track. (a) What is x? (b) If the block were to reach the top of the track, what would be its speed at that point? (c) Does the block actually reach the top of the track, or does it fall off before reaching the top?

Short answer

(a) The value of x is $\mathrm{x}=0.400\mathrm{m}$.

(b) The speed of the block at the top point is${\mathrm{v}}_{\mathrm{f}}=4.10\mathrm{m}/\mathrm{s}$ .

(c) The block stays on the track.

Step-by-step solution

Step 1:

If non-conservative forces act on a body, the total change in the mechanical energy is equal to the work done against the non-conservative forces.

This can be written as:

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}\mathbf{0}$

Here:

$\mathrm{\Delta U}=$Change in Potential energy

$\mathrm{\Delta K}=$Change in kinetic energy

${\mathrm{\Delta E}}_{\mathrm{int}}=$Change in internal energy

Step 2

Part (a):

Using the law of conservation of energy for an isolated spring-block system, we have:

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ {(\mathrm{K}+\mathrm{U})}_{\mathrm{i}}={(\mathrm{K}+\mathrm{U})}_{\mathrm{f}}\\ \left(\frac{1}{2}{\mathrm{mv}}^2-0\right)+\left(0-\frac{1}{2}{\mathrm{kx}}^2\right)=0\\ \mathrm{x}=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\mathrm{v}.\end{array}$

Here, there is a change in the kinetic energy of the system, which is a change in the potential energy of the system. The subscripts $\left(i\right)$and $\left(f\right)$denote the initial and the final values of the quantities respectively.

For the given values, the above equation becomes as shown below.

$\begin{array}{l}\mathrm{x}=\sqrt{\frac{0.500\text{kg}}{450\text{N/m}}}(12.0\text{m/s})\\ =0.400\text{m.}\end{array}$

Part (b):

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ {(\mathrm{K}+\mathrm{U})}_{\mathrm{i}}={(\mathrm{K}+\mathrm{U})}_{\mathrm{f}}\\ \left(\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2-\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2\right)+(2\mathrm{mgR}-0)+{\mathrm{f}}_{\mathrm{k}}\left(\mathrm{\pi R}\right)=0\\ {\mathrm{v}}_{\mathrm{f}}=\sqrt{{\mathrm{v}}_{\mathrm{i}}^2-4\mathrm{gR}-\frac{2{\mathrm{\pi f}}_{\mathrm{k}}\mathrm{R}}{\mathrm{m}}}\end{array}$

For the given values:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{f}}=\sqrt{{(12.0\text{m/s})}^2-4(9.8{\text{m/s}}^{\text{2}})(1.00\text{m})-\frac{2\mathrm{\pi }(7.00\text{N})(1.00\text{m})}{0.500\text{kg}}}\\ =4.10\text{m/s.}\end{array}$

Part(c):

If${\mathrm{a}}_{\mathrm{c}}$ is the net acceleration acting on the body, ${\mathrm{a}}_{\mathrm{c}}<\mathrm{g}$, and the block will fall from the track before reaching the top point.

$\begin{array}{c}{\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}_{\mathrm{T}}^2}{\mathrm{R}}\\ =\frac{{(4.10\text{m/s})}^2}{1.00\text{m}}\\ =16.8{\text{m/s}}^2\end{array}$

As ${\mathrm{a}}_{\mathrm{c}}>\mathrm{g}$(acceleration due to gravity), the block stays on the track and reaches the top point.