Suspending two blocks connected by a string of negligible mass over a pulley. the blocks are initially held at rest and then released at time t = 0 find the speed of the block at t = 4?

  v = (m₁ –m₂) / (m₁ + m₂) 39.2

Explanation:

For this exercise, let's use Newton's second law, since the mass of the chain is zero has forces. Let's call the block on the left 1 and the one on the right 2

For the block on the left

                 W₁ - T = m₁ a

For the block on the right

               T - W₂ = m₂ a

Where the acceleration (a) is the same for the entire system, so that the chain does not lose tension,

To solve this system of equations let's add it

               W₁ - W₂ = (m₁ + m₂) a

               a = (m₁ –m₂) / (m₁ + m₂) g

Now we can use the kinematic relationship to find the speed of the block

              v = v₀ + a t

              v = 0 + (m₁ –m₂) / (m₁ + m₂) g t

We substitute the time t = 4 s

               v = (m₁ –m₂) / (m₁ + m₂) 9.8 4

               v = (m₁ –m₂) / (m₁ + m₂) 39.2

For a specific value we must have the mass of the blocks

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6