# Spring constant problem: Work done, speed of block, maximum compression

A 230 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.5 N/cm (Fig. 7-30). The block becomes attached to the spring and compresses the spring 14 cm before momentarily stopping.

What work is done on the block by the spring force while the spring is being compressed? (in J)

What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (in m/s)

If the speed at impact is doubled, what is the maximum compression of the spring? (in m)

## Solution This solution is **FREE** courtesy of BrainMass!

What you have here is an interesting problem in conservation of energy. What makes springs interesting is that the force required to stretch the spring is directly proportional to the distance it is stretched. Mathematically this is given by Hooke's Law, which states that the F=kx, where k is the "spring constant", a measure of how tight the spring is. Springs with high spring constants are very "tight" (consider the spring in your pen). In contrast springs like slinkys are very loose, meaning their spring constant is small. In your case, k=2.5 N/cm, meaning that you would have to supply 2.5 N of force to stretch or compress your spring 1 cm.

The first part of your problem asks "How much work is done on the block by the spring force?" The definition of work on an object by a force is the product of the amount of force acting on that object times the distance the force acts. For example if I pull on a box with 10 N of force over a distance of 5 m, I will do 50 N*m = 50 Joules of work on the box. According the Newton's third law the force exerted by the box on the spring is equal to the force exerted by the spring on the box. We then know that as the spring is compressed a distance x, it exerts a force k*x (Hooke's law) on the box. Thus, one may believe that the work done on the box is simply F*x = k*x*x=k*x^2. This is incorrect in this case! Note that the force the spring exerts on the box is not always the same (for your spring if the compression is 1cm the force is 2.5 N, 2 cm the force is 5 N etc...). You need to add up all of the force contributions at each distance to get the total work. This requires calculus, and if you work it out, the work done is actually 1/2k*x*2. Thus the work done on the box is simply 1/2*spring constant*(distance the box compresses the spring)^2. Note to have proper units (J) you need to convert all masses to kg, all distances to m, and the spring constant to N/m (thus your k becomes 250 N/m)

The question I have is where does all the energy from the box go? As we see above, it goes into compressing the spring. Indeed, if the mass is removed, the spring will "shoot up" from its compressed position and release all the energy that was given to it. Thus the formula 1/2*k*x^2 is doubly useful. Not only does it tell you how much work the spring did on the box; it also tells you how much energy is stored in the spring! This is potential energy; the energy stored in an object due to its position. Thus, for a spring, its potential energy is 1/2*k*x^2

The question then becomes, what kind of energy does the block have before it hits the spring. Objects that are falling can have two kinds of energy:

1) Kinetic energy (KE) (energy due to velocity). The formula for KE=1/2mv^2

2) Potential Energy (PE), which is the energy stored in the block as it rises of the ground. For example, you know if you launch an object in the air its velocity slows as it rises. The object eventually stops and then speeds up again upon its return to earth. From an energy perspective, the object has traded kinetic energy for potential energy as it rose, and then exchanged the PE for KE again when it returned to earth. The formula for PE=m*g*h, where m is the mass of the object, g is the accel. Due to gravity (typically 9.81 m/s^2) and h is the height of the object above the ground.

Since energy is conserved when dealing with gravity or springs, here is the key point we need to solve the second part of your problem:

The total amount of energy stored in the box + spring before the collision = The total amount of energy stored in the box + spring after the collision.

If we can set up an equation that describes this accurately, we will know both answers to the last parts of the problem.

How much energy does the box have when it just hits the spring? Well

1) It has kinetic energy = 1/2*m*v^2 (that v is what we are looking for in Q2).

2) It also has potential energy m*g*h. The question is, what is h? In this problem we are going to assume h=0 whenever the spring is compressed its maximum amount. The great thing about potential energy due to falling is you get to define what zero height is. Since the minimum height of the box is when sits on the compressed spring, we will define that as 0. Thus, in your problem, the PE of the box is m*g*h=m*g*x=.230*9.81*.14 (note we know that the box is 14 cm above its min. height when it hits. Thus the height of the box is just the distance the spring will compress when the box is stopped)

What kind of energy does the spring have before the collision? Well, since it is sitting there, uncompressed, it has nothing. So its energy is 0.

Thus the total energy of the mass and spring at the instant before the collision is:

1/2mbox*(vel box)^2 + mbox*g*.14 (I will let you insert the rest of your numbers. Remember kg and m as units!)

What about after the collision? Well the box is stopped with no velocity and its height, by our definition, is 0. Thus the box has no KE or PE. For the spring, its only energy now is the energy it stored in the collision with the box. From your first question, that is equal to 1/2*k*x^2 = .5*250*.14^2. This is the only contributor to energy after the collision.

We are now ready to solve the problem. Since:

Energy before = Energy After

1/2mbox*vbox^2 + mbox*g*ht of box + 0 = 0 + 1/2k*x^2

Note the first 0 is the initial energy of the spring. The zero after the equals sign is the final energy of the box. When you solve for v, I get an answer around 4.1-4.5 m/s (I will let you get the actual answer)

To solve the final problem, note that the physics is the same and the same equation must be valid. However you now know the velocity of the box, and what you need to solve for is x (the distance the spring compresses). Thus, we will get

1/2m*(2v)^2 + m*g*x = 1/2*k*x^2.

where (2v) is twice the velocity you solved for in the previous problem. Note you will have to solve a quadratic equation here, using the quadratic formula x=(-b+/-SQRT(b^2-4*a*c))/2a . When you do this, you will get two x's. One will be negative and of course can't be the answer to the problem. The pos. x value will probably lie between .18 m and .23 m, which makes sense because the box had more energy to give the spring.

The general equation you have above should always help in problems like this. Try to use energy conservation whenever possible, because it makes problems that would be really hard to do using Newton's Laws a lot easier