Three blocks are connected on the table as shown below. The coefficient of kinet

Question

Three blocks are connected on the table as shown below. The coefficient of kinetic friction between the block of mass m2 and the table is 0.305. The objects have masses of m1 = 3.25 kg, m2 = 1.35 kg, and m3 = 2.50 kg, and the pulleys are frictionless

Determine the acceleration of each object, including its direction. m1: magnitude 2 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2
direction 3 ---Select--- up down .
m2: magnitude 4 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2
direction 5 ---Select--- left right .
m3: magnitude 6 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2
direction 7 ---Select--- up down .

Explanation / Answer

A for m1 is down for m2 is to the left and m3 is up Free body diagrams about each yields m1*g - Tleft = m1*a so T (left) = m1*g - m1*a T (left) - T (right) - µ*m2*g = m2*a T (right) - m3*g = m3*a So T (right) = m3*g + m3*a Now sub the first and third into the second m1*g - m1*a - m3*a - m3*g - µ*m2*g = m2a So (m1 + m2 + m3)*a = g*(m1 - m3 - µ*m2) So a = g*(m1- m3 - µ*m2) /(m1 + m2 + m3) = 9.8*(3.25 - 2.50 - 0.305*1.35)/(3.25 +1.35 + 2.5) = 0.466 m/s^2 The directions were given above Now T (left) = m1*g - m1*a = 3.25*(9.8 - 0.466) = 30.33N T (right) = m3*g + m3*a = 2.50*(9.8 + 0.466) = 23.33N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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