Chapter 7, Problem 58AP

(a)

To determine

The forces exerted on the block by the tracks at points A and B.

Answer to Problem 58AP

Solution:

The force exerted by the track on the block at point A and B is is respectively $15\text{N}$ and $30\text{N}$.

Explanation of Solution

Given info: The mass of the block is $0.50\text{kg}$. The initial speed of the block is $4.0\text{m}\cdot {\text{s}}^{\text{-1}}$. The radius of the circular track is $1.5\text{m}$.

From work-energy theorem,

$W_n=\left(K{E}_f+P{E}_f\right)-\left(K{E}_i+P{E}_i\right)$

• $W_n$ is the work done by the non-conservative forces
• $K{E}_f$ is the final kinetic energy
• $P{E}_f$ is the final potential energy
• $K{E}_i$ is the initial kinetic energy
• $P{E}_i$ is the initial potential energy

The kinetic energy of an object is given by,

$KE=\frac{1}{2}m{v}^2$

• $m$ is the mass
• $v$ is the velocity

In the given case the potential energy is the gravitational potential energy. Thus,

$PE=mgy$

• $g$ is the free fall acceleration
• $y$ is the height of the block from the zero potential level

Consider the level of point B as the zero potential state. The point A is the initial position.

Substitute the expression for the kinetic and potential energy with point C as the initial state, the work-energy theorem can be written as,

$\begin{array}{c}{W}_n=\left(\frac{1}{2}m{v}^2+mgy\right)-\left(\frac{1}{2}m{v}_0^2+2mgR\right)\\ =\frac{1}{2}m\left(v^2-v_0^2\right)+mg\left(y-2R\right)\end{array}$

• $v_0$ is the initial velocity of the block
• $R$ is the radius of the circular track

Re-write the above equation in order to get an expression for the velocity,

$v^2=v_0^2+\frac{2W_n}{m}+2g\left(2R-y\right)$

At the point A, the track is frictionless. So there is no non-conservative force at a point A. Moreover at point A $y=R$. Thus the above expression can be written as,

$\begin{array}{c}{v}^2=v_0^2+2g\left(2R-R\right)\\ =v_0^2+2gR\end{array}$

The force acting on the block by the track is the centripetal force. Thus,

$n=\frac{m{v}^2}{R}$

Rewrite the above equation using $v_0^2+2gR$ for v.

$n=\frac{m\left(v_0^2+2gR\right)}{R}$

Substitute $0.50\text{kg}$ for $m$, $4.0\text{m}\cdot {\text{s}}^{\text{-1}}$ for $v_0$, $1.5\text{m}$ for $R$ and $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$ find n.

$\begin{array}{c}n=\frac{\left(0.50\text{kg}\right)\left({\left(4.0\text{m}\cdot {\text{s}}^{\text{-1}}\right)}^2+2\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left(1.5\text{m}\right)\right)}{\left(1.5\text{m}\right)}\\ =15\text{N}\end{array}$

From work-energy theorem,

$W_n=\left(K{E}_f+P{E}_f\right)-\left(K{E}_i+P{E}_i\right)$

• $W_n$ is the work done by the non-conservative forces
• $K{E}_f$ is the final kinetic energy
• $P{E}_f$ is the final potential energy
• $K{E}_i$ is the initial kinetic energy
• $P{E}_i$ is the initial potential energy

The kinetic energy of an object is given by,

$KE=\frac{1}{2}m{v}^2$

• $m$ is the mass
• $v$ is the velocity

In the given case the potential energy is the gravitational potential energy. Thus,

$PE=mgy$

• $g$ is the free fall acceleration
• $y$ is the height of the block from the zero potential level

Consider the level of point B as the zero potential state. The point A is the initial position.

Substitute the expression for the kinetic and potential energy with point C as the initial state, the work-energy theorem can be written as,

$\begin{array}{c}{W}_n=\left(\frac{1}{2}m{v}^2+mgy\right)-\left(\frac{1}{2}m{v}_0^2+2mgR\right)\\ =\frac{1}{2}m\left(v^2-v_0^2\right)+mg\left(y-2R\right)\end{array}$

• $v_0$ is the initial velocity of the block
• $R$ is the radius of the circular track

Re-write the above equation in order to get an expression for the velocity,

$v^2=v_0^2+\frac{2W_n}{m}+2g\left(2R-y\right)$

At the point B, the track is still frictionless. So there is no non-conservative force at a point A. And at point B, $y=0$. Thus the above expression can be written as,

$\begin{array}{c}{v}^2=v_0^2+2g\left(2R-0\right)\\ =v_0^2+4gR\end{array}$

The force acting on the block by the track at point B is the sum of centripetal force and the weight of the block. Thus,

$n=\frac{m{v}^2}{R}+mg$

Substituting the expression for the velocity,

$\begin{array}{c}n=m\left(\frac{\left(v_0^2+4gR\right)}{R}+g\right)\\ =m\left(\frac{\left(v_0^2+5gR\right)}{R}\right)\end{array}$

Substitute $0.50\text{kg}$ for $m$, $4.0\text{m}\cdot {\text{s}}^{\text{-1}}$ for $v_0$, $1.5\text{m}$ for $R$ and $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$ to determine the force exerted on the block at point A.

$\begin{array}{c}n=\left(0.50\text{kg}\right)\left(\frac{\left({\left(4.0\text{m}\cdot {\text{s}}^{\text{-1}}\right)}^2+5\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left(1.5\text{m}\right)\right)}{\left(1.5\text{m}\right)}\right)\\ =30\text{N}\end{array}$

Thus, the force exerted by the track on the block at point B is $30\text{N}$.

Conclusion:

The force exerted by the track on the block at point A and B is $15\text{N}$ and $30\text{N}$ respectively.

(b)

To determine

The coefficient of friction between the block and the portion of the track where friction is present.

Answer to Problem 58AP

Solution:

The coefficient of friction between the block and the track where friction is present is $0.17$.

Explanation of Solution

Given info:

The mass of the block is $0.50\text{kg}$. The initial speed of the block is $4.0\text{m}\cdot {\text{s}}^{\text{-1}}$. The radius of the circular track is $1.5\text{m}$. The length of the track where friction is present is $0.40\text{m}$.

From work-energy theorem,

$W_n=\left(K{E}_f+P{E}_f\right)-\left(K{E}_i+P{E}_i\right)$

• $W_n$ is the work done by the non-conservative forces
• $K{E}_f$ is the final kinetic energy
• $P{E}_f$ is the final potential energy
• $K{E}_i$ is the initial kinetic energy
• $P{E}_i$ is the initial potential energy

The kinetic energy of an object is given by,

$KE=\frac{1}{2}m{v}^2$

• $m$ is the mass
• $v$ is the velocity

In the given case the potential energy is the gravitational potential energy. Thus,

$PE=mgy$

• $g$ is the free fall acceleration
• $y$ is the height of the block from the zero potential level

Consider the level of point B as the zero potential state. The point A is the initial position.

Substitute the expression for the kinetic and potential energy with point C as the initial state, the work-energy theorem can be written as,

$\begin{array}{c}{W}_n=\left(\frac{1}{2}m{v}^2+mgy\right)-\left(\frac{1}{2}m{v}_0^2+2mgR\right)\\ =\frac{1}{2}m\left(v^2-v_0^2\right)+mg\left(y-2R\right)\end{array}$

• $v_0$ is the initial velocity of the block
• $R$ is the radius of the circular track

Re-write the above equation in order to get an expression for the velocity,

$v^2=v_0^2+\frac{2W_n}{m}+2g\left(2R-y\right)$

Consider the block at point C. At point C the height of the block with respect to the zero potential $y$ is $2R$. Thus the above expression becomes,

$\begin{array}{c}{v}^2=v_0^2+\frac{2W_n}{m}+2g\left(2R-2R\right)\\ =v_0^2+\frac{2W_n}{m}\end{array}$ (I)

The work done by the non-conservative forces will be equal to the work done by the frictional force.

$W_n=-F_{\text{Friction}}L$

• $F_{\text{Friction}}$ is the frictional force
• $L$ is the length of the track where friction is present

But the frictional force is defined as the coefficient of friction time the normal force,

$F_{\text{friction}}=\mu mg$

• $\mu$ is the coefficient of friction

Substitute the expression for the frictional force,

$W_n=-\mu mgL$ (II)

Substitute (II) in (I),

$\begin{array}{c}{v}^2=v_0^2+\frac{2\left(-\mu mgL\right)}{m}\\ =v_0^2-2\mu gL\end{array}$ (III)

More over at point C the centripetal force is given by the weight of the block.

$\frac{m{v}^2}{R}=mg$

Re-write the above equation for an expression for $v$,

$v^2=Rg$ (IV)

Equate the right hand sides of (III) and (IV),

$Rg=v_0^2-2\mu gL$

Re-write the above equation for an expression for $\mu$,

$\mu =\frac{v_0^2-Rg}{2gL}$

Substitute $0.50\text{kg}$ for $m$, $4.0\text{m}\cdot {\text{s}}^{\text{-1}}$ for $v_0$, $1.5\text{m}$ for $R$ and $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$, $0.40\text{m}$ for $L$  determine $\mu$,

$\begin{array}{c}\mu =\frac{{\left(4.0\text{m}\cdot {\text{s}}^{\text{-1}}\right)}^2-\left(1.5\text{m}\right)\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)}{2\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left(0.40\text{m}\right)}\\ =0.17\end{array}$

Conclusion:

The coefficient of friction between the block and the track where friction is present is $0.17$.