Spring work and velocity with friction and frictionless surfaces

[Hadou]

Okay...so I have Physics homework due at midnight, 8 hours. And I'm stuck on a few problems...posting here for some help.
An explanation of how to do it would be much more appreciated then just answer.

1a)A 1816 g mass is on a horizontal surface with μk = 0.48, and is in contact with a massless spring with a force constant of 610 N/m which is compressed. When the spring is released, it does 1.952 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed. Figured this out to be just 1.952 = .5kx^2. Before I was factoring in friction.
b)What is the velocity of the mass as it loses contact with the spring?

2a)A 0.65 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 850 N/m) whose other end is fixed. The mass has a kinetic energy of 17.0 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?
b)At what rate is the spring doing work on the mass when the spring is compressed 0.100 m and the mass is moving away from the equilibrium position?

I'm lost with springs.


Thanks...

 

[member 553083]

1a) To solve this problem, we first need to find the initial compression of the spring. This can be done using the equation W = ½kx², where W is the work done by the spring, k is the force constant of the spring, and x is the initial compression of the spring. Substituting in the given values, we get 1.952 = ½(610)x². Solving for x gives us a value of 0.094 meters. b) The velocity of the mass when it loses contact with the spring can be found using the equation v = √(2W/m), where v is velocity, W is the work done by the spring, and m is the mass. Substituting in the given values gives us v = √(2(1.952)/1816), which simplifies to v = 0.0644 m/s. 2a) To solve this problem, we first need to find the energy stored in the spring. This can be done using the equation E = ½kx², where E is the energy stored in the spring, k is the force constant of the spring, and x is the compression of the spring. Since the mass is at equilibrium, the compression of the spring is zero, so we have E = ½k(0)², which simplifies to E = 0. Thus, the rate at which the spring is doing work on the mass as it passes through its equilibrium position is 0.b) Now, we need to find the rate at which the spring is doing work on the mass when the spring is compressed 0.100 m and the mass is moving away from the equilibrium position. We can do this using the equation W = ½kx², where W is the work done by the spring, k is the force constant of the spring, and x is the compression of the spring. Substituting in the given values, we get W = ½(850)(0.100)², which simplifies to W = 0.425 J. Thus, the rate at which the spring is doing work on the mass is 0.425 J/s.

 

[wais]

First of all, it's great that you are seeking help and not just looking for someone to give you the answers. Understanding the concepts and being able to solve the problems on your own is key in learning physics.

For the first problem, you are on the right track. The work done by the spring is equal to the change in potential energy of the mass. So, using the equation you mentioned, 1.952 = 0.5kx^2, you can solve for x which will give you the distance the spring was compressed.

For part b), you can use the conservation of energy principle to solve for the velocity of the mass. The initial kinetic energy of the mass will be equal to the final potential energy of the spring. So, you can set up the equation 0.5mv^2 = 0.5kx^2 and solve for v.

Moving on to the second problem, for part a) the spring is doing work on the mass at a constant rate as it passes through its equilibrium position because the spring force is zero at that point. So, the rate of work done by the spring will be equal to the kinetic energy of the mass divided by the time it takes to pass through its equilibrium position.

For part b), the spring is compressed and the mass is moving away from the equilibrium position, so the work done by the spring will be negative. You can use the same equation as in part a) but with the new distance the spring is compressed to calculate the rate of work done by the spring.

Remember to always pay attention to the units and make sure they are consistent throughout the calculations. Also, try to draw diagrams and visualize the situation to better understand the problem.

I hope this helps and good luck with your homework!