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Physics / Physics with MasteringPhysics 4 / Chapter 8 / Problem 95GP

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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Consider the system shown in Figure. (a) What initial

Chapter 8, Problem 95GP

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QUESTION:

IP Consider the system shown in Figure 8-31. (a) What initial speed $v$ is required if the blocks \(m_1=2.4\ \mathrm {kg}\) and \(m_2=1.1\ \mathrm {kg}\) are to travel a distance \(d=6.5\ \mathrm {cm}\) before coming to rest? Assume the coefficient of kinetic friction between \(m_1\) and the tabletop is \(\mu_\mathrm k=0.25\). (b) Is the work done on \(m_2\) by the rope positive, negative, or zero? Explain. (c) Calculate the work done on \(m_2\) by the rope.

Equation Transcription:

${m}_{1}=2.4\ kg$

${m}_{2}=1.1\ kg$

$d=6.5\ cm$

${m}_{1}$

${\mu{}}_{k}=0.25$

${m}_{2}$

${m}_{2}$

${m}_{1}$

${m}_{2}$

Text Transcription:

m_1=2.4 kg

m_2=1.1 kg

d=6.5 cm

m_1

k=0.25

m_2

m_2

m_1

m_2

Questions & Answers

QUESTION:

IP Consider the system shown in Figure 8-31. (a) What initial speed $v$ is required if the blocks \(m_1=2.4\ \mathrm {kg}\) and \(m_2=1.1\ \mathrm {kg}\) are to travel a distance \(d=6.5\ \mathrm {cm}\) before coming to rest? Assume the coefficient of kinetic friction between \(m_1\) and the tabletop is \(\mu_\mathrm k=0.25\). (b) Is the work done on \(m_2\) by the rope positive, negative, or zero? Explain. (c) Calculate the work done on \(m_2\) by the rope.

Equation Transcription:

${m}_{1}=2.4\ kg$

${m}_{2}=1.1\ kg$

$d=6.5\ cm$

${m}_{1}$

${\mu{}}_{k}=0.25$

${m}_{2}$

${m}_{2}$

${m}_{1}$

${m}_{2}$

Text Transcription:

m_1=2.4 kg

m_2=1.1 kg

d=6.5 cm

m_1

k=0.25

m_2

m_2

m_1

m_2

ANSWER:

Solution 95GP

Step 1 of 3

Part a

We are required to calculate the initial speed $v$ .

Total initial energy of the system,

$E{\ }_{i}=\frac{1}{2}(m{\ }_{1}+m{\ }_{2}){v}^{2}$

Total final energy of the system,

$E{\ }_{f}=m{\ }_{2}gd$

Therefore, the net work done is,

$W=m{\ }_{2}gd-\frac{1}{2}(m{\ }_{1}+m{\ }_{2}){v}^{2}$ …..(1)

The coefficient of kinetic friction between the block of mass $m{\ }_{1}$ and the tabletop is $\mu{}{\ }_{k}$ .

So, the work done to displace the block $m{\ }_{1}$ is,

$W=\ -\mu{}{\ }_{k}m{\ }_{1}gd$ …..(2) ( negative sign is used to indicate the work done against friction)

Equating equations (1) and (2),

$-\mu{}{\ }_{k}m{\ }_{1}gd=m{\ }_{2}gd-\frac{1}{2}(m{\ }_{1}+m{\ }_{2}){v}^{2}$

$(\mu{}{\ }_{k}m{\ }_{1}+m{\ }_{2})gd=\frac{1}{2}(m{\ }_{1}+m{\ }_{2}){v}^{2}$ …..(3)

Given data:

$m{\ }_{1}=2.4$ kg

$m{\ }_{2}=1.1$ kg

$d=6.5$ cm

$d=0.065$ m

$\mu{}{\ }_{k}=0.25$

Substitute in equation (3),

$(0.25\times{}2.4+1.1)\times{}9.81\times{}0.065=\frac{1}{2}\times{}(2.4+1.1)\times{}{v}^{2}$

$v=0.79$ m/s

The required speed $v$ is 0.79 m/s.

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