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Harlamova29_29 [7]
3 years ago
7

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

le is 0.228. What is the maximum value of m2 before m1 starts sliding across the table?
Physics
1 answer:
slamgirl [31]3 years ago
5 0

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_F between the mass m1 and the table is given as follows;

F_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_F = 2.4525 N × 0.228 = 0.55917 N

F_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_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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Answer:

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Explanation:

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distance of electron from rod 1 = r₂ = 0.4 m

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Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

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Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

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