Suppose that a 150g mass(0.15kg) ocillates at the end of a spring on a friction free hotizontal surface.?

Question

The spring can be both stretched and compressed and has a spring constant of 250 N m. It was originally stretched a distance of 10 mc(0.1m) from its equilibreium position before release.

A. What is its initial kinetic energy?

B. What is the maximun velocity that the mass will reach its ocillation? Where in the motion is the maximum released?

C. Ignoring friction, what are th valuse of the potential energy, kinetic energy, and velocity of the mass when it is 5cm from the equlibrium position.

D. How does the value of velovity computed in part C compare to part B? (What is the ratio of the values?)

Answer 1

Now I am sure you know about Hooke's law which is

F=K*(Xi-Xo)

F = Force

K = spring constant

Xi= the springs equilibrium position

Xo= the position the spring has been compressed or stretched

Now that doesn't help with finding the amount of energy that can be stored in a spring, but wait we have our good friend calculus to help us turn that force into energy.

Also according to the work energy theorem

Work = Total Energy

which the total energy = Potential + kinetic

and work is equal to the dot product of force into distance

W=F(dot)x

The total work done on the system can be found using a line integral  and for integral I will use int(A) where A will be the integrated item, and since we are doing an integral x will become dx.

W=Int(F(dot)dx)  

Now back to Hooke's Law.

F=K(Xi-Xo) since we know that if we dot dx into F and integrate we will get it into work and you can drop the Xi and Xo and replace it with just x since Xi and Xo will the limits for integration.

int(F(dot)dx)=int(K*x(dot)dx) but we can drop the dot products and just replace them with multiplication since the dot product of two parallel vectors is equal to the magnitude of the two vectors multiplied into each other.

so this becomes

int(F*dx)=int(K*x*dx) with the upper and lower limits being Xo and Xi. Also int(F*dx)= Work which I will use W.

We know have W=(1/2)*K*(Xi-Xo)^2

Also remeber the work energy theorem

W = Energy

and also Total energy = Potential energy + kinetic energy.

Remember Kinetic energy = (1/2)*mass*velocity^2

use this to find the velocity when it asks for it.

So now that the easy work is out of the way onto the hard.

Now for part A) since the spring was stretched to the final position we can assume that the initial conditions for velocity will be zero so the kinetic energy will be zero

Ok part B is next using the fact that the total energy in the system will be equal to the work and they are looking for the maximum velocity just plug the numbers into W=(1/2)*(Xi-Xo)^2 and remember total= kinetic+potential and potential in this instance will be zero, I would also assume Xi to be zero makes life easier to place the initial value at the origin, and also remeber that the negative sign will mean it is going left positive means it will be going right.

C) that is easy, you know the total energy in the system, see part B, so what you need to do is find out how much is in potential and how much is in the kinetic.

Take that trusty equation at the end of the lecture and instead of using the 0.1 meters plug in the 0.05 meters. Now assume you streched it to that point, so set the Kinetic energy equal to zero and that will tell you how much of the energy is potential . Now go back to that equation Total Energy=Kinetic + Potential

and plug in the W for when the spring was streched

0.1 meters

and for the potential energy place the potential energy of the spring streched 0.05 meters. and solve for solve for kinetic.

D) is easy once you work on B and C but a hint will be the velocity of C should be less then B