Question

The equation below results from the application of a physical principle to a physical system

– (0.25)(3kg)(9.8m/s2)(1m) = 0.5(3kg)v2 + 0.5(5kg)v2 - (5kg)(9.8m/s2)(1m)

(a) [10pts] Sketch and label a situation that would result in this equation

(b) [10pts] Explain how your sketch is consistent with the equation. Be sure to include your potential energy reference(s).

Answer 1

The velocity of the system is 3.24 meters per second.

Step-by-step explanation:

The given expression is

`-(0.25)(3 kg)(9.8 m/s^(2))(1m)=0.5(3kg)v^(2) +0.5(5kg)v^(2) -(5kg)(9.8m/s^(2))(1m)`

The physical principle used here is the conservation of energy.

In this case, we have a physical system formed by two bodies, one of them with mass of 3 kilograms and the other with mass of 5 kilograms.

We know that the energy cannot be created or destroyed, it can be transformed.

In the given expression, we can observe that the net energy of the system is equal to zero
`K+U=0`, then
`-U=K` which is the given expression.

In the physical system, the 3 kilogram mass is at 1 meter from the level ground and the 5 kilogram mass is also at this level, so basically their potential energies are different because their masses are different.

However, the system conservates in a way that both velocities are the same.

So, using the conservation of energy theorem and all given values, we can find the velocity of the system.

`-6.96=1.5v^(2)+2.5v^(2)-49\\ -6.96+49=4v^(2)\\v^(2)=(42.04)/(4) \\ v^(2) =10.51\\v=√(10.51) \approx 3.24`

Therefore, the velocity of the system is 3.24 meters per second.

To answer the second question, our sketch is consistent because it follows the conservation of energy theorem, which states that the net energy of a system is constant:
`\Delta K +\Delta U =0`