Question

Computer the initial velocity of a 100g (mass) projectile that is launched by stretching a rubber band by 10cm. Assume the rubber band has an elastic constant of 50 KN/m and obeys a F = 4KS law. Hints:Potential (elastic) energy = FS (force x distance stretched) = KS x S Initial Velocity = Square root of[(2 x energy )/ mass]

A. 0.1 m/s
B. 1 m/s
C. 10 m/s
D. 100 m/s
E. No correct answer is available

Answer 1

Again, let's use the work energy theorem. I've already derived this for you in your previous question, so I'll jump to the point. \[PE_s = {1 \over 2} 4 kx^2 = 2 kx^2\]The '4' present here is because the force of the spring is 4kx, not the usual kx.

From conservation of energy\[KE_p = PE_s\]where \(KE_p\) is the kinetic energy of the projectile. \[KE_p = {1 \over 2} mv^2\]

Therefore, \[{1 \over 2} mv^2 = 2 kx^2\]

Answer 2

I honestly do not know any of this stuff. Its on my engineering take home test and we havent studied any formulas. This is over my head.

I dont even think he meant to put this in the test.

Answer 3

Those formulas are way out of my league. I havent studied this before. (yet)

Answer 4

This is our 3rd test and none of the previous test have involved math. Usually just definitions from the book.

Like I said. I really dont think this belongs in the test.

Answer 5

Really? Is this your first physics course? This is Physics 101 material.

Regardless, this solution assumes an understanding of energy. Let me explain further.

Energy, in a mechanical sense (there are several types of energy), is represented in two forms. 1) Kinetic energy - the energy of motion and 2) Potential energy - energy of position.

The kinetic energy of a body of mass m with a velocity v is given as\[KE = {1 \over 2} mv^2\]

The potential energy of a body of mass m, that h meters about the surface of the earth (relative to the surface of the earth) is given as\[PE = mgh\]

Now, in the case of a spring, realize that we are not necessarily changing the height, but instead compressing the spring. By compressing the spring, we apply a force over some distance. If we compress the spring, we store energy in it. This, as previously given, is represented as\[PE_s = {1 \over 2} kx^2\]where k is the spring constant and x the distance the spring has been compressed.

Understand so far?

Answer 6

All I need is a basic engineering class for my computer science major. Physics isn't required so I haven't been introduced to this.

But yea, I think I understand that. So far.

Answer 7

What university do you attend?

Now, the principle of conservation of energy states that energy can neither be created nor destroyed, it can only be converted into another form.

In this problem, we will say that all the potential energy stored in the spring will be converted to kinetic energy.

That is to say\[PE_s = KE_p \rightarrow {1 \over 2} kx^2 = {1 \over 2} mv^2\]

However, in this problem, the energy of the spring is not \[{1 \over 2} kx^2\]because this expression is derived from a spring force of \(F_s = kx\). Here \(F_s = 4 kx\), therefore, we have the following expression for conservation of energy\[2 kx^2 = {1 \over 2} mv^2\]You can easily solve this equation for v. Keep in mind k has units of Newtons/meter, x has units of meters, m has units of kg, and v has units of meters/second.