The speed of block B in terms of the distance (d) it has descended is ![V=\frac{2gd(m_B-\mu_k m_A)}{m_A \;+\; m_B \;+\;\frac{I}{r^2}}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B2gd%28m_B-%5Cmu_k%20m_A%29%7D%7Bm_A%20%5C%3B%2B%5C%3B%20m_B%20%5C%3B%2B%5C%3B%5Cfrac%7BI%7D%7Br%5E2%7D%7D)
<h3>How to determine the speed.</h3>
In order to determine the speed of block B in terms of the distance (d) it has descended, we would apply the work-energy theorem as follows:
![Work=\Delta KE=KE_f-KE_i](https://tex.z-dn.net/?f=Work%3D%5CDelta%20KE%3DKE_f-KE_i)
The work done due to gravity is given by:
![W_g=(m_B)gd](https://tex.z-dn.net/?f=W_g%3D%28m_B%29gd)
The work done due to friction is given by:
![W_f=-\mu_k (m_A)gd](https://tex.z-dn.net/?f=W_f%3D-%5Cmu_k%20%28m_A%29gd)
The angular velocity of pulley is given by:
![\omega = \frac{V}{r}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7BV%7D%7Br%7D)
Applying the work-energy theorem, we have:
![W_g+W_f= KE_f-KE_i\\\\(m_B)gd-\mu_k (m_A)gd=\frac{1}{2} m_A V^2+\frac{1}{2} m_B V^2+\frac{1}{2} I\omega^2-0\\\\gd(m_B-\mu_k m_A)=m_A V^2+ m_B V^2+\frac{1}{2} I(\frac{V}{r} )^2\\\\2gd(m_B-\mu_k m_A)=(m_A + m_B +\frac{I}{r^2} )V\\\\V=\frac{2gd(m_B-\mu_k m_A)}{m_A \;+\; m_B \;+\;\frac{I}{r^2}}](https://tex.z-dn.net/?f=W_g%2BW_f%3D%20KE_f-KE_i%5C%5C%5C%5C%28m_B%29gd-%5Cmu_k%20%28m_A%29gd%3D%5Cfrac%7B1%7D%7B2%7D%20m_A%20V%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m_B%20V%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20I%5Comega%5E2-0%5C%5C%5C%5Cgd%28m_B-%5Cmu_k%20m_A%29%3Dm_A%20V%5E2%2B%20m_B%20V%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20I%28%5Cfrac%7BV%7D%7Br%7D%20%29%5E2%5C%5C%5C%5C2gd%28m_B-%5Cmu_k%20m_A%29%3D%28m_A%20%2B%20m_B%20%2B%5Cfrac%7BI%7D%7Br%5E2%7D%20%29V%5C%5C%5C%5CV%3D%5Cfrac%7B2gd%28m_B-%5Cmu_k%20m_A%29%7D%7Bm_A%20%5C%3B%2B%5C%3B%20m_B%20%5C%3B%2B%5C%3B%5Cfrac%7BI%7D%7Br%5E2%7D%7D)
Read more on work-energy here: brainly.com/question/22599382
Gatling gun this is the answer