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 tab
1 answer:
Answer:
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.
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