Help finding potential energy of spring.

[jgibbon2]

Block A has mass 1.00 kg, and block B has mass 3.00 kg. The blocks are forced together, compressing a spring between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.10 m/s.

a) What is the final speed of block A? Found this answer to be 3.30 m/s

b) How much potential energy was stored in the compressed spring?
- how do you find this answer?? KEa+KEb=9.075 ... not right

 

[PhanthomJay]

check again your math in the KE calculation.

 

[m2003]

To find the potential energy stored in the compressed spring, we can use the formula for potential energy of a spring: PE = 1/2*k*x^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position.

In this scenario, the spring is compressed and then released, so the displacement x is equal to the initial compression of the spring. We can use the formula for Hooke's law, F = -kx, to find the spring constant k.

Since the system is released from rest and there is no friction, we can use the principle of conservation of energy to equate the potential energy stored in the spring to the kinetic energy gained by the blocks.

Therefore, we have the equation: 1/2*k*x^2 = 1/2*mA*vA^2 + 1/2*mB*vB^2, where mA and mB are the masses of blocks A and B, and vA and vB are their final velocities.

Using the given values, we can solve for the spring constant k: k = -F/x = -(mA + mB)*g/x = -4*9.8/0.05 = -784 N/m.

Substituting this value of k into the equation, we get: PE = 1/2*(-784)*(0.05)^2 = 1.96 J.

Therefore, the potential energy stored in the compressed spring is 1.96 J.

 

[ogb p]

To find the potential energy stored in the compressed spring, you can use the formula for potential energy of a spring: PE = 1/2*k*x^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this scenario, the spring is compressed, so x would be the distance that the spring is compressed.

First, you need to find the spring constant (k). This can be done by using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. So, you can set up the equation F = kx and use the given information to solve for k. In this case, the force exerted by the spring is equal to the weight of block B, which is 3.00 kg * 9.8 m/s^2 = 29.4 N. The displacement of the spring can be calculated by using the final speed of block B and the time it takes to reach that speed. Since block B has a speed of 1.10 m/s and starts from rest, the average speed would be 0.55 m/s. The time can be found using the formula t = d/v, where d is the distance the spring compresses and v is the average speed. In this case, d can be calculated by using the equation for the displacement of an object under constant acceleration, d = 1/2*a*t^2, where a is the acceleration and t is the time. The acceleration can be found using the formula a = (vf-vi)/t, where vf is the final speed and vi is the initial speed (which is 0 in this case). So, a = (1.10 m/s - 0 m/s)/t = 1.10 m/s^2. Plugging in the values, we get t = 0.5 s. Now, we can calculate d = 1/2*(1.10 m/s^2)*(0.5 s)^2 = 0.1375 m. Therefore, x = 0.1375 m.

Now, we can plug in the values of k and x into the formula for potential energy, PE = 1/2*k*x^2. We already found k to be 29.4 N/m, so PE = 1/2*(29.4 N/m)*(0.1375 m)^2 =