Question

A first class lever is balanced with 25 kg suspended from one side and 1 kg suspended from the other. 1 extra kg is placed on the 25 kg side. What is the velocity of the 1 kg after the 26 kg has dropped .3 m?

Answer 1

To determine the velocity of the 1 kg mass after the 26 kg mass has dropped by 0.3 m in a first-class lever system, we need to consider the principle of conservation of momentum.

In a first-class lever, the torque produced by the masses on either side of the fulcrum must be balanced for equilibrium. The torque is calculated as the product of the mass, the acceleration due to gravity, and the distance from the fulcrum.

Initially, when the system is balanced with a 25 kg mass on one side and a 1 kg mass on the other side, the torques on both sides are equal. After adding an extra 1 kg mass to the 25 kg side, the torques become unbalanced, causing the system to move.

Let's assume that the 1 kg mass moves downward by a distance of 0.3 m. To find the velocity of the 1 kg mass after this displacement, we can apply the principle of conservation of momentum.

The change in momentum of the system can be calculated by multiplying the mass difference between the two sides (26 kg - 1 kg = 25 kg) by the change in velocity of the center of mass. Since the center of mass is lowered by 0.3 m, we can calculate the change in velocity ($\Delta v$) using the equation:

$\Delta v=\frac{\Delta h}{t}$

Where: $\Delta h$ is the change in height (0.3 m) and $t$ is the time taken for the displacement.

Since we don't have information about the time taken for the displacement, we cannot directly determine the change in velocity. However, we can determine the final velocity of the 1 kg mass once we have the time.

If you provide the time taken for the 0.3 m displacement, I can calculate the final velocity of the 1 kg mass.