I am confident I am on the right track... please help me. The 17-kg block A and

Question

I am confident I am on the right track... please help me.

The 17-kg block A and 27-kg cylinder B are connected by a light cord that passes over a 5-kg pulley (disk). (Figure 1) If the system is released from rest, determine the cylinder's velocity after its has traveled downwards 2 m. The coefficient of kinetic friction between the block and the horizontal plane is micro k = 0.27. Assume the cord does not slip over the pulley. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

m_c*a = m_c*g - T

m_b*a = T - F

Here T is tension in string. F is friction force = *normal force

Normal force = m_b*g = 17*9.81 = 166.77 N

F = 0.27*166.77 = 45.0279 N

Adding first two eqns, (m_b + m_c)*a = m_c*g - F

(17+27)*a = 27*9.81 - 45.0279

a = 4.996 m/s^2

Now using, v^2 = u^2 + 2as we get

v^2 = 0 + 2*4.996*2

v = 4.47 m/s (We are not given friction at pulley and hence assuming that pulley is smooth.)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.