We'll use the Newton's second law:

$\overline{){\mathbf{\Sigma}}{\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

**Part (a)**

The force acting on the block m_{1} in the x-direction is T.

We have, T = m_{1}a_{1}

The acceleration is:

a_{1} = T/m_{1}

The acceleration that the block of mass m_{1} experiences in the x-direction is **T/m**_{1}.

**Part (b)**

Block of mass m_{2} experiences T in the positive y-direction and m_{2}g in the negative y-direction.

We have the sum of forces as:

m_{2}g - T = m_{2}a_{2}

Two blocks are connected by a massless rope. The rope passes over an ideal (frictionless and massless) pulley such that one block with mass m_{1} = 15 kg is on a horizontal table and the other block with mass m_{2} = 5.5 kg hangs vertically. Both blocks experience gravity and the tension force, T. Use the coordinate system specified in the diagram

Part (a) Assuming friction forces are negligible, write an expression, using only the variables provided, for the acceleration that the bloc of mass m experiences in the x-direction. Your answer should involve the tension, T

Part (b) Under the same assumptions, write an expression for the acceleration, a_{2}, the block of mass m_{2} experiences in the y-direction Your answer should be in terms of the tension, T and m_{2}.

Part (c) Carefully consider how the accelerations a_{1} and a_{2} are related. Solve for the magnitude of the acceleration, a_{1}, of the block of 250 mass m_{1}, in meters per square second.

Part (d) Find the magnitude of the tension in the rope, T, in newtons.

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