Question

An object of mass m1 = 4.90 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m2 = 7.40 kg as shown in the figure.

Answer #1

The acceleration and tension for both blocks will be the same ignoring friction and the moment of inertia of the pulley. From Newton's 2nd law, write equations for the forces acting on each block separately :

ΣF(m₁) = m₁a = T (T, tension, is the only force acting on the block since the table is frictionless).

ΣF(m₂) = m₂a = m₂g - T

Adding these two equations together gives:

m₁a + m₂a = m₂g

Solving for a :

a = m₂g / (m₁+ m₂)

= 7.4kg(9.80m/s²) / (4.9kg + 7.4kg)

= **5.9
m/s²**

From m₁a = T, we get :

T = 4.9kg(5.9m/s²)

= **28.9 N**

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