7. A mass m1 of 250 g is on a table connected to a massless pulley, as shown. The coefficient of friction between m1 and the table is 0.228. What is the maximum value of m2 before m1 starts sliding across the table?

The minimum mass of m2 before m1 starts sliding across the table is 57 grams

Explanation:

The given mass on the table, m1 = 250 g

The coefficient of friction between the table and the mass, m1, μ = 0.228

Therefore, the friction force, F between the mass m1 and the table is given as follows;

F = W₁ × μ

Where;

W₁ = The weight of m1  = Mass of m1 × Acceleration due to gravity

W₁ = 250 g × 9.81 m/s² = 0.25 kg × 9.81 m/s² = 2.4525 N

W₁ = 2.4525 N

Which gives;

F = 2.4525 N × 0.228 = 0.55917 N

F = 0.55917 N

By equilibrium of forces, the weight, W₂, of the mass m2 at which the mass m1 starts sliding is equal to the frictional force F of m1 or 0.55917 N

Therefore, for slipping to occur, we have;

W₂ = 0.55917 N

However, W₂ = Mass of m2 × Acceleration due to gravity

∴ W₂ = 0.55917 N = m2 × 9.81 m/s²

0.55917 N = m2 × 9.81 m/s²

m2 = 0.55917 N/(9.81 m/s²) = 0.057 kg = 57 grams

The minimum mass of m2 before m1 starts sliding across the table is 57 grams.

f=ma newton

f=100x2.5=250n