Kundan who weighs 200N and bijaya who weighs 500N are playing sea saw. If Bijaya is 1m away from the fulcrum, how far should kundan sit to lift bijaya

Kundan should sit at more that 2.5 meters from the location of the fulcrum on the side opposite to the side where Bijaya sits

Explanation:

The given parameters are;

The weight of Kundan, F₁ = 200 N

The weight of Bijaya, F₂ = 500 N

The distance of Bijaya from the fulcrum, d₂ = 1 m

Therefore, for Kundan to lift Bijaya, Kundan should sit at a location further than the point where both the moment of Bijaya and Kundan about the fulcrum are equal

Moment of a force, M = The magnitude of the force, F × The perpendicular distance to the point at which the force acts, d

At equilibrium, the sum of the clockwise moment = The sum of the anticlockwise moment

Taking the moment of Kundan as being in the anticlockwise direction and the moment of Bijaya as acting in the clockwise direction, we have;

F₁ × d₁ = F₂ × d₂

∴ 200 N × d₁ = 500 N × 1 m

d₁ = 500 N·m/(200 N) = 2.5 m

Therefore, when Kundan sits at the point d₁ = 2.5 m, away from the fulcrum on one side of the sea saw and Bijaya sits at the point 1 m from the fulcrum on the other side of the sea saw, both Kundan and Bijaya will have the same moment

For Kundan to lift Bijaya, therefore, Kundan as to sit at a point further than 2.5 m from the fulcrum on the opposite side of the fulcrum from Bijaya.

answer: a ball

explanation: because it will bounce

it has to be b, about 97 percent.