A block of mass M1 sits on a table, with no friction, and ablock of mass M2 is s
ID: 1672652 • Letter: A
Question
A block of mass M1 sits on a table, with no friction, and ablock of mass M2 is suspended by a cord of negligible mass over africtionless pulley. At t=0s the blocks are at rest, and thehanging mass starts descend.
Student X says: "The hanging mass is pulling down on the rope,so the tension in the rope is (M2). That's the force on theblock on the table, so that block's acceleration isa=F/m=(M2)g/(M1)."
Student Y says: "That can't be right, because then if M2 isbigger than M1, then M1 will have an acceleration bigger thang!"
Find a better expression for the acceleration of M1.
I am confused about what they are really asking me todescribe.
Explanation / Answer
For the block on the table: T = Fnet = m1a For the hanging block Fnet = T + m2g assuming downwardacceleration T = m2a - m2g m1a = m2a - m2g m2g = m2a + m1a m2g = (m2 + m1)a a = m2g/(m2 + m1) Hope that helpsRelated Questions
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