Question

As shown in Figure P8.20, a green bead of mass 25 g slides along a straight wire. The length of the wire from point A to point B is 0.600 m, and point A is 0.200 m higher than point B. A constant friction force of magnitude 0.0250 N acts on the bead. (a) If the bead is released from rest at point A, what is its speed at point B? (b) A red bead of mass 25 g slides along a curved wire, subject to a friction force with the same constant magnitude as that on the green bead. If the green and red beads are released simultaneously from rest at point A, which bead reaches point

B with a higher speed? Explain.

Short answer

(a)The speed at point B is$v_B=1.65\mathrm{m}/\mathrm{s}$$v_B=1.65\mathrm{m}/\mathrm{s}$

Step-by-step solution

Step 1:

Considering the book moves on an incline that is not frictionless, there is a change in both the kinetic energy and the gravitational potential energy of the book Earth system.

$$\begin{gathered} \mathbf{∆}{\mathbf{E}}_{\mathbf{mech}}\mathbf{=}\mathbf{∆}\mathbf{K}\mathbf{+}\mathbf{∆}{\mathbf{U}}_{\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}{\mathbf{f}}_{\mathbf{k}}\mathbf{d} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{∆}{\mathbf{E}}_{\mathbf{int}} \end{gathered}$$

This change in mechanical energy is equal to the increase in internal energy of the book. In general, if a non-conservative force acts as an isolated system,

role="math" localid="1668156305620" $\mathbf{∆}\mathbf{K}\mathbf{+}\mathbf{∆}\mathbf{U}\mathbf{+}\mathbf{∆}{\mathbf{E}}_{\mathbf{int}}\mathbf{=}\mathbf{0}$

Where,

$\mathbf{∆}\mathbf{K}\mathbf{=}$Change in kinetic energy

$\mathbf{∆}\mathbf{U}\mathbf{=}$Change in potential energy

$\mathbf{∆}{\mathbf{E}}_{\mathbf{int}}\mathbf{=}$Change in internal energy

Given Data

Mass of beads: m = 25 g

Length of wire: L = 0.60 m

Vertical separation between points: d = 0.20 m

Frictional force: F = 0.250 N

Solving Part (a)

Using the law of conservation of energy in the bead-string-Earth system.

$∆K+∆U+∆E_{int}=0$

As internal energy change occurs due to friction,

$∆E_{int}=f_{k}d$

Using above equations, we get-

$\left(k_B-k_A\right)+\left(u_B-u_A\right)+f_{k}d=0$

Substitute the initial and final energies:

$$\begin{gathered} \left(\frac{1}{2}m{v}_B^2-\frac{1}{2}m{v}_A^2\right)+\left(mg{y}_B-mg{y}_A\right)+f_{k}d=0 \\ \left(\frac{1}{2}m{v}_B^2-0\right)+\left(0-mg{y}_A\right)+f_{k}d=0 \\ \frac{1}{2}m{v}_B^2=mg{y}_A-f_{k}d \end{gathered}$$

Thus, velocity at point B will be-

$$\begin{gathered} v_B=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.200\mathrm{m}\right)-\frac{2\left(0.025\mathrm{N}\right)\left(0.600\mathrm{m}\right)}{25\times {10}^{-3}\mathrm{kg}}} \\ =\sqrt{2.72}\mathrm{m}/\mathrm{s} \\ =1.65\mathrm{m}/\mathrm{s} \end{gathered}$$

The velocity of ball at B will be 1.65 m/s.