Chapter 8, Problem 8.65AP

A block of mass 0.500 kg is pushed against a horizon-tal 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 Ⓐ, 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 v_Ⓐ = 12.0 m/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?

To determine

The value of compression in the spring $x$.

Answer to Problem 8.65AP

The value of compression in the spring $x$ is $0.400\text{m}$.

Explanation of Solution

Given info: The mass of the block is $0.500\text{kg}$, value of spring force constant is $450\text{N}/\text{m}$, radius of the track is $1.00\text{}\text{m}$, speed of the block at point $A$ is $12.0\text{m}/\text{s}$, average friction force is $7.00\text{}\text{N}$.

The formula to calculate the initial kinetic energy of the block is,

$K_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2$

Here,

$m$ is the mass of the block.

$v_{\text{i}}$ is the initial velocity of the block.

The initial velocity of the block is 0 as it is at rest then the initial kinetic energy of the block is 0.

The formula to calculate the final kinetic energy is,

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2$

Here,

$m$ is the mass of the block.

$v_{\text{f}}$ is the final velocity of the block.

The formula to calculate initial potential energy is,

$U_{\text{i}}=\frac{1}{2}k{x}^2$

Here,

$k$ is the spring constant.

$x$ is the initial compression distance.

The formula to calculate the final potential energy is,

$U_{\text{f}}=\frac{1}{2}k{x}_{\text{f}}{}^2$

Here,

$k$ is the spring constant.

$x_{\text{f}}$ is the final compression distance.

Thus, the final potential energy of the block is $\frac{1}{2}k{x}_{\text{f}}{}^2$.

The formula to calculate the initial energy is,

$E_{\text{i}}=U_{\text{i}}+K_{\text{i}}$

Here,

$U_{\text{i}}$ is the initial potential energy.

$K_{\text{i}}$ is the initial kinetic energy.

The final compression distance is 0 as the spring does not move after striking to the block then the final potential energy is 0.

Substitute $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K_{\text{i}}$ and $\frac{1}{2}k{x}^2$ for $U_{\text{i}}$ in the above formula to find $E_{\text{i}}$.

$E_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}^2$

Thus, the initial energy is $\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}^2$.

The formula to calculate the final energy is,

$E_{\text{f}}=U_{\text{f}}+K_{\text{f}}$

Here,

$U_{\text{f}}$ is the final potential energy.

$K_{\text{f}}$ is the final kinetic energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_{\text{f}}$ and $\frac{1}{2}k{x}_{\text{f}}{}^2$ for $U_{\text{f}}$ in the above formula to find $E_{\text{f}}$.

$E_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$

Thus, the final energy is $\frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$.

From the law of conservation of the energy,

$E_{\text{f}}=E_{\text{i}}$

Here,

$E_{\text{f}}$ is the final energy.

$E_{\text{i}}$ is the initial energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}^2+\frac{1}{2}k{x}_{\text{f}}{}^2$ for $E_{\text{f}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}^2$ for $E_{\text{i}}$ in the above formula to find $x$.

$\begin{array}{c}{E}_{\text{f}}=E_{\text{i}}\\ \frac{1}{2}m{v}_{\text{f}}{}^2+\frac{1}{2}k{x}_{\text{f}}{}^2=\frac{1}{2}m{v}_{\text{i}}{}^2+\frac{1}{2}k{x}^2\end{array}$

Substitute $0$ for $\frac{1}{2}m{v}_{\text{i}}{}^2$ and $0$ for $\frac{1}{2}k{x}_{\text{f}}{}^2$ in the above formula to find $x$ .

$0+\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_{\text{f}}{}^2+0$

Rearrange the above formula for $x$.

$\begin{array}{c}\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_{\text{f}}{}^2\\ x^2=\frac{m{v}_{\text{f}}{}^2}{k}\\ x=\sqrt{\frac{m{v}_{\text{f}}{}^2}{k}}\end{array}$

Substitute $0.500\text{}\text{kg}$ for $m$ , $12.0\text{m}/\text{s}$ for $v_{\text{f}}$ , $450\text{N}/\text{m}$ in the above formula to find $x$.

$\begin{array}{c}x=\sqrt{\frac{m{v}_2{}^2}{k}}\\ =\sqrt{\frac{\left(0.500\text{kg}\right){\left(12.0\text{m}/\text{s}\right)}^2}{\left(450\text{N}/\text{m}\right)}}\\ =0.400\text{m}\end{array}$

Conclusion:

Therefore, the value of $x$ is $0.400\text{m}$.

To determine

The speed of the block at the top of the track.

Answer to Problem 8.65AP

The speed of the block at the top of the track is $4.10\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the block is $0.500\text{kg}$ , value of spring force constant is $450\text{N}/\text{m}$, radius of the track is $1.00\text{}\text{m}$, speed of the block at point $A$ is $12.0\text{m}/\text{s}$, average friction force is $7.00\text{}\text{N}$.

The formula to calculate the work done by the frictional force is,

$W=-F\cdot x$

Here,

$F$ is the frictional force.

$x$ is the displacement.

The formula to calculate the initial kinetic energy of the block is,

$K_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2$

Here,

$m$ is the mass of the block.

$v_{\text{i}}$ is the initial velocity of the block.

The formula to calculate the final kinetic energy is,

$K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2$

Here,

$m$ is the mass of the block.

$v_{\text{f}}$ is the final velocity of the block.

The formula to calculate initial potential energy is,

$U_{\text{i}}=mg{h}_{\text{i}}$

Here,

$m$ is the mass of the block.

$g$ is the acceleration due to gravity.

$h_{\text{i}}$ is the initial height of the block.

The initial height of the block is 0 as the block is at the bottom of the track then the initial potential energy is 0.

The formula to calculate the final potential energy is,

$U_{\text{f}}=mg{h}_{\text{f}}$

Here,

$m$ is the mass of the block.

$g$ is the acceleration due to gravity.

$h_{\text{f}}$ is the final height of the block.

Substitute $2R$ for $h_{\text{f}}$ in the above formula to find $U_{\text{f}}$.

$\begin{array}{c}{U}_{\text{f}}=mg{h}_{\text{f}}\\ =mg\left(2R\right)\\ =2Rmg\end{array}$

Thus, the final potential energy of the block is $2Rmg$.

The formula to calculate the initial energy is,

$E_{\text{i}}=U_{\text{i}}+K_{\text{i}}$

Here,

$U_{\text{i}}$ is the initial potential energy.

$K_{\text{i}}$ is the initial kinetic energy.

Substitute $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K_{\text{i}}$ and $mg{h}_{\text{i}}$ for $U_{\text{i}}$ in the above formula to find $E_{\text{i}}$.

$E_{\text{i}}=\frac{1}{2}m{v}_{\text{i}}{}^2+mg{h}_{\text{i}}$

Thus, the initial energy is $\frac{1}{2}m{v}_1{}^2+mg{h}_1$.

The formula to calculate the final energy is,

$E_{\text{f}}=U_{\text{f}}+K_{\text{f}}$

Here,

$U_{\text{f}}$ is the final potential energy.

$K_{\text{f}}$ is the final kinetic energy.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2$ for $K_{\text{f}}$ and $2mgR$ for $U_{\text{f}}$ in the above formula to find $E_{\text{f}}$.

$E_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}{}^2+2mgR$

Thus, the final energy is $\frac{1}{2}m{v}_{\text{f}}{}^2+2mgR$.

The formula to calculate the law of conservation of energy is,

$F_{\text{f}}-E_{\text{i}}=W$

Here,

$E_{\text{f}}$ is the final energy.

$E_{\text{i}}$ is the initial energy.

$W$ is the work done by the frictional force.

Substitute $\frac{1}{2}m{v}_{\text{f}}{}^2+mg{h}_{\text{f}}$ for $E_{\text{f}}$ , $\frac{1}{2}m{v}_{\text{i}}{}^2+mg{h}_{\text{i}}$ for $E_{\text{i}}$ and $-Fx$ for $W$ in the above formula to find $v_{\text{f}}$.

$\begin{array}{l}{F}_{\text{f}}-E_{\text{i}}=-Fx\\ \frac{1}{2}m{v}_{\text{f}}{}^2+mg{h}_{\text{f}}-\left(\frac{1}{2}m{v}_{\text{i}}{}^2+mg{h}_{\text{i}}\right)=-Fx\end{array}$

Substitute $0$ for $mg{h}_1$ and $2mgR$ for $mg{h}_2$ in the above formula to find $v_{\text{f}}$ .

$\begin{array}{l}\frac{1}{2}m{v}_{\text{f}}{}^2+mg{h}_{\text{f}}-\left(\frac{1}{2}m{v}_{\text{i}}{}^2+mg{h}_{\text{i}}\right)=-Fx\\ \frac{1}{2}m{v}_{\text{f}}{}^2+2mgR-\frac{1}{2}m{v}_{\text{i}}{}^2=-Fx\end{array}$

Rearrange the above formula for $v_{\text{f}}$

$\begin{array}{c}\frac{1}{2}m{v}_{\text{f}}{}^2+2mgR-\frac{1}{2}m{v}_{\text{i}}{}^2=-Fx\\ v_{\text{f}}=\frac{2\left(-Fx+\frac{1}{2}m{v}_{\text{i}}{}^2-2mgR\right)}{m}\\ v_{\text{f}}=\sqrt{\frac{2\left(-Fx+\frac{1}{2}m{v}_{\text{i}}{}^2-2mgR\right)}{m}}\end{array}$

Substitute $0.500\text{}\text{kg}$ for $m$ , $12.0\text{m}/\text{s}$ for $v_{\text{f}}$ , $1.00\text{m}$, $7.00\text{N}$ for $F$, $9.8{\text{m}/\text{s}}^2$ for $g$  in the above formula to find $x$.

$\begin{array}{c}{v}_{\text{f}}=\sqrt{\frac{2\left(-Fx+\frac{1}{2}m{v}_{\text{i}}{}^2-2mgR\right)}{m}}\\ =\sqrt{\frac{\begin{array}{l}2\left(-7.00\text{N}\right)\left(3.14\right)\left(1.00\text{m}\right)+\frac{1}{2}\left(0.500\text{kg}\right)\left(12.0\text{m}/\text{s}\right)-\\ \left(2\left(0.500\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(1.00\text{m}\right)\right)\end{array}}{\left(0.500\text{kg}\right)}}\\ =\sqrt{\frac{4.21\text{J}}{0.25\text{kg}}}\\ =4.10\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the block at the top of the track is $4.10\text{}\text{m}/\text{s}$.

To determine

Whether the block reach the top of the track or fall off before reaching the top.

Answer to Problem 8.65AP

The block stays at the top of the track.

Explanation of Solution

Given info: The mass of the block is $0.500\text{kg}$, value of spring force constant is $450\text{N}/\text{m}$, radius of the track is $1.00\text{}\text{m}$, speed of the block at point $A$ is $12.0\text{m}/\text{s}$, average friction force is $7.00\text{}\text{N}$.

The formula to calculate the centripetal acceleration of the block at the top of the track is,

$a_{\text{c}}=\frac{v_{\text{f}}{}^2}{R}$

Here,

$R$ is the radius of the track.

$v_2$ is the velocity of the block at the top of the track.

Substitute $4.1\text{}\text{m}/\text{s}$ for $v_{\text{f}}$ and $1.00\text{m}$ for $R$ in the above formula to find $a_{\text{c}}$.

$\begin{array}{c}{a}_{\text{c}}=\frac{v_{\text{f}}{}^2}{R}\\ =\frac{{\left(4.1\text{}\text{m}/\text{s}\right)}^2}{\left(1.00\text{m}\right)}\\ =16.81{\text{m}/\text{s}}^2\end{array}$

if the centripetal acceleration of the block at the top of the track is less than the acceleration due to gravity then the block fall but centripetal acceleration of the block at the top of the track is greater than the acceleration due to gravity that concludes that the block actually reaches the top of the track.

Conclusion:

Therefore, the block stays at the top of the track.