What is drop is potential energy if M1 goes down by d1∆θ and M2 goe

1.	What is drop is potential energy if M1 goes down by d1∆θ and M2 goes up by d2∆θ in case of two bodies of masses M1 and M2 sitting on a seesaw and balancing each other initially?
Answer» If M₁ goes down by d₁∆θ and M₂ goes up by d₂∆θ then the drop in potential energy is, ∆U = M₂gd₂∆θ - M₁gd₁∆θ ∆U = (F₂d₂ - F₁d₁)∆θ ∵ Mg₂ = F₂ and M₁g = F₁

What is drop is potential energy if M1 goes down by d1∆θ and M2 goes up by d2∆θ in case of two bodies of masses M1 and M2 sitting on a seesaw and balancing each other initially?

Answer»

If M₁ goes down by d₁∆θ and M₂ goes up by d₂∆θ then the drop in potential energy is,

∆U = M₂gd₂∆θ - M₁gd₁∆θ

∆U = (F₂d₂ - F₁d₁)∆θ

∵ Mg₂ = F₂ and M₁g = F₁