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)© BrainMass Inc. brainmass.com June 22, 2018, 7:33 am ad1c9bdddf
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 ...
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