Dynamics question (kinetics and energy)

[gomerpyle]

Homework Statement

The two blocks are connected by a light inextensible cord, which passes around small

massless pulleys as shown below. If block B is pulled down 500 mm from the equilibrium

position and released from rest, determine its speed when it returns to the equilibrium

position.

http://s3.amazonaws.com/answer-board...7387504353.gif

Homework Equations

T1 + U1 = T2 + U2

The Attempt at a Solution

If it's pulled down below equilibrium and held there, then T1 of the system is zero because both blocks are not moving. At the moment B passes through the equilibrium, there is no more potential energy, only kinetic, then the equation would look like:

U1 = T2

The problem I'm running into is that I get a negative value for the left side of this equation, which is impossible because then it would have to go under a square root when solving for the velocity.

for U1 I had:

mgha - mghb + 0.5kx^2

Since 'b' moves down 0.5m, a moves up 0.25 and the spring is stretched 0.25. Is this right since A is attached to the pulley and B is simply hanging? If that's the case then:

(2)(9.81)(0.25) - (10)(9.81)(0.5) + 0.5(800)(0.25)^2

Which is negative. Supposedly the answer is 2.16 m/s.

 

[KataKoniK]

Thank you for your post. Your approach seems to be on the right track, however there are a few things that need to be clarified.

Firstly, when you say "If it's pulled down below equilibrium and held there, then T1 of the system is zero because both blocks are not moving," this is not entirely accurate. The tension in the cord will still be present, as it is the force that is keeping block B in place. The tension T1 will be equal to the weight of block B, which is 10 kg * 9.81 m/s^2 = 98.1 N.

Secondly, when considering the potential energy of the system, it is important to define a reference point for the potential energy. In this case, it is helpful to choose the equilibrium position of block B as the reference point, as this is where the potential energy will be zero. This means that the potential energy at point A will be equal to the potential energy at point B, as they are both at the same height above the reference point. Therefore, your equation for potential energy should be:

U1 = U2

mgha = mghb

Thirdly, when solving for the velocity at the equilibrium position, you can use the conservation of energy equation:

U1 + K1 = U2 + K2

Since the block is released from rest, K1 = 0. Therefore:

U1 = U2 + K2

Substituting in the equations for potential energy and kinetic energy:

mgha = mghb + 0.5mv^2

Solving for v:

v = √(2gha - 2ghb)

Finally, plugging in the given values:

v = √(2 * 9.81 m/s^2 * 0.25 m - 2 * 9.81 m/s^2 * 0.5 m)

v = √(2.45 m/s^2 - 9.81 m/s^2)

v = √(-7.36 m/s^2)

v = 2.16 m/s (rounded to two significant figures)

I hope this helps clarify the solution. If you have any further questions or concerns, please don't hesitate to ask.Scientist