A 200 gram mass is placed on the meter stick 20 cm from the pivot point (the fulcrum). An unknown mass is positioned 8 cm from the fulcrum to balance the system. What is the mass of this unknown object?

Answer :

Answer:

80 grams

Explanation:

multiplying 8 by 2.5 to get 20

so then deviding 200 by that 2.5 will get you 80 grams

This question involves the concepts of equlibrium and moment.

The mass of unknown object should be "500 g".

EQUILIBRIUM

In order to balance the meter stick equilibrium should be achieved. For equilibrium the moment due to each mass must be the same.

[tex]F_1d_1=F_2d_2\\m_1gd_1=m_2gd_2\\m_1d_1=m_2d_2[/tex]

where,

  • m₁ = mass of known object = 200 g
  • m₂ = mass of unknown object = ?
  • d₁ = distance from fulcrum of known object = 20 cm
  • d₂ = distance from fulcrum of unknown object = 8 cm

Therefore,

[tex](200\ g)(20\ cm)=(m_2)(8\ cm)\\\\m_2=\frac{(200\ g)(20\ cm)}{8\ cm}[/tex]

m₂ = 500 g

Learn more about moment here:

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