Please explain detailed with detailed answer step by step. Thank you! When a 0.2
ID: 1441507 • Letter: P
Question
Please explain detailed with detailed answer step by step. Thank you! When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.40 m to 0.45 m. The same block is attached to the same spring and placed on a horizontal, frictionless surface as shown. The block is then pulled so that the spring stretches to a total length of 0.15 m. The block is released at time t = 0 s and undergoes simple harmonic motion. What is the speed of the block each time the spring is 0.50 m long?
Explanation / Answer
Hi
In this case you must remember some concepts of dynamics (such as elastic force and the second law) as well as some notions of energy conservation.
In order to answer the question at hand, we can use the Theorem of Energy's conservation, considering the block and the spring as a system:
To do that we must consider two points of the movement of the block while it lies horizontally in its harmonic motion.
The first one is when it is stretched for the first time. In that instant all the mechanical energy in the system is concentrated in the form of elastic potential energy.
The second one is when the the spring is a 0.5 m long. In that moment the mechanical energy of system is divided between the potential (elastic one) and the kinetic energy.
As the block is put over a frictionless surface, the movement continues forever and the energy of those two points mentioned above must be equal, as the total energy of the system is conserved.
Expressing this in a mathematical way, we have the following:
(1/2) kx02 = (1/2) kd2+ (1/2) mv2 ; where m is the mass of the block, k is the constant of the spring, v is the speed of the block, d is the value of the total distance (0.5 m) minus the original lenght of the spring and x0 is the difference between the distance 0.15 m and the original lenght of the spring.
The previous expression could be used to find the speed of the block, but we don't know the value of the constant of the spring.
However, we can find said value using dynamics. The first part of the exercise tells us that the block hangs vertically and its weight is countered by the force of the spring. With the data they provide us for that situation, we can find the constant of the spring:
If the block is at rest, then the elastic force and the weight of the block must have equal value, so:
Fs = Fw :::::: xk = mg ; where g is the acceleration due to gravity and x is the total length of spring minus the length of its initial position, so:
k = mg/x = (0.2 kg)*(9.8 m/s2) /(0.45 m - 0.4 m) = 39.2 N/m
Using this value, the problem is solved:
v = [ k/m ( x02 - d2) ]1/2 = [ (39.2 N/m / 0.2 kg)*( (0.15 m - 0.4 m)2 - (0.5 m - 0.4 m)2 ]1/2 = 3.21 m/s
I hope it helps.