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Two blocks are suspended from opposite ends of a light rope that passes over a light, frictionless pulley. One block has mass m1 and the other has mass m2, where m2>m1.The two blocks are released from rest, and the block with mass m2 moves downward 6.00 m in 2.00 s after being released. While the blocks are moving, the tension in the rope is 18.0 N.

Part A

Calculate m1.

Express your answer with the appropriate units.

Part B

Calculate m2.

Express your answer with the appropriate units.

Respuesta :

Answer:

a)  m₁ = 1.41 kg , b)  m₂ = 2.65 kg

Explanation:

For this exercise we will use Newton's second law

Block 1

             T - W₁ = m₁ a

Block 2

           W₂ - T = m₂ a

We have selected the positive  block 1 rising and block two lowering, as the pulley has no friction does not affect the movement

Let's use kinematics to look for acceleration

         y = v₀ t + ½ a t²

As part of the rest the initial speed is zero

           a = 2 y / t²

           a = 2 6.00 / 2²

           a = 3 m / s²

Let's replace in the equation of block 1

a)   T = m₁ g + m₁ a

     m₁ = T / (g + a)

     m₁ = 18.0 / (9.8 + 3)

     m₁ = 1.41 kg

b) we substitute in the equation of block 2

     W₂ - T = m₂ a

     m₂ g - m₂ a = T

     m₂ = T / (g-a)

     m₂ = 18.0 / (9.8 -3)

    m₂ = 2.65 kg

okpalawalter8

The value for M1 and M2  is mathematically given as

m1 = 1.41 kg

m2 = 2.65 kg

Mass of Block

Generally the equation for Newton's law is mathematically given as

s=ut+1/2at^2

Therefore

a = 2 6.00 / 2²

a=3m/s

Hence

a)

T = m₁ g + m₁ a

m1 = 18.0 / (9.8 + 3)

m1 = 1.41 kg

b) for block 2

W₂ - T = m₂ a

m2 = 18.0 / (9.8 -3)

m2 = 2.65 kg

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