You are given a set of blocks. Block one is on the top of the table,

Question

You are given a set of blocks. Block one is on the top of the table, it is connected to block two

using a string and a frictionless pully. Block two is hanging down along the side of the table.

Block one has a mass of 0.3 Kg and block two has a mass of 0.5 Kg. If the coefficient of friction is

0.05 between block one and the table, determine the acceleration of the system. Draw your

picture for full credit.

Answer 1

Please draw a picture of the situation.

Let

M₁ = 0.3 kg be the mass of the block on the table,

M₂ = 0.5 kg be the mass of the block hanging down,

μ = 0.05 be the coefficient of friction,

g = 9.81 m/s² be gravitational acceleration,

a (to be computed) be the acceleration of the system,

T (unknown) be the tension on the string.

The statement of the problem does not mention the mass of the pully.

Therefore I will assume that it has mass 0.

That implies that both bocks experience the same tension T on the string.

Assuming that the string does not stretch, the two blocks experience the same acceleration a.

The block hanging down experiences two forces:

- a downward force of gravity of magnitude M₂g

- an upward force of tension of magnitude T

The block on the table experiences four forces:

- a downward force of gravity of magnitude M₁g

- an upward normal force of magnitude also M₁g

- the force of tension directed towards the pully of magnitude T

- the force of friction directed away from the pully of magnitude M₁gμ.

The net force acting on the box on table is directed towards the pully

and has magnitude T - M₁gμ.

By Newton's Second Law

M₁a = T - M₁gμ (1)

The net force acting on the box hanging down is directed downward

and has magnitude M₂g - T.

By Newton's Second Law

M₂a = M₂g - T (2)

Adding equations (1) and (2)

M₁a + M₂a = M₂g - M₁gμ

From that

a = g(M₂ - M₁μ)/(M₁ + M₂)

Plugging in actual numbers

a = 9.81×(0.5 - 0.3×0.05)/(0.3 +0.5) ≈ 6 m/s²