Question

A spring with a mass attached to it is stretched from the rest position of 20 cm to the position of 87 cm. If the spring constant of this spring is 1729 N/m, what is the potential energy stored in this spring?
561.29 J
120.97 J
388.07 J
254.79 J

Answer 1

Answer:

388.07 J

Explanation:

Answer 2

Answer:

C) 388.07

Explanation:

It is a spring therefore, there must be Elastic potential energy. Now what is the formula for elastic potential energy? It is 1/2 *k *x^2

K=spring constant

x= displacement  

1/2=.5 (hate fractions)

Now, Displacement simply means (how much was it moved?) So, that means subtract.

87-20= 67   x=67

Next, do x^2 (so you don't make a mistake by squaring everything :)  )

67*67=4,489

NOW, do....

.5*1729*4,489=3,880,740.5

Now convert to Joules, I just did 3,880,740.5  / 10^3 = 3880.745

round to 3880.7 J  

                                                                ~There you go loves !