Question

please help!

In 1992, David Engwall of California used a slingshot to launch a dart with a mass of 62 g. The dart traveled a horizontal distance of 477 m. Suppose the slingshot had a spring constant of 3.0 x 10^4 N/m. If the elastic potential energy stored in the slingshot just before the dart was launched was 1.4 x 10^2 , how far was the slingshot stretched?

Answer 1

We want to find how far was the slingshot stretched. We will see that it was stretched 10 centimeters approx.

Working with springs.

Remember that for a spring of constant K, the elastic potential energy is given by:

$P = \frac{1}{2}K*X^2$

Where X is how far the spring is stretched.

Now, we can think of a slingshot as a spring, we know that the constant is:

• K = 3.0x 10^4 N/m

And that the elastic potential energy is:

• P =  1.4x10^2 N*m

Now we can just replace these two in the above equation to find the value of X:

$P = \frac{1}{2}K*X^2\\\\X = \sqrt{2*P/K} = \sqrt{2*(1.4*10^2 Nm)/(3.0* 10^4 N/m)} = 0.1 m$

So the slingshot was stretched 10cm approx.

If you want to learn more about springs, you can read:

https://brainly.com/question/2901244