Suppose that 5 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 36 cm.
(a) How much work is needed to stretch the spring from 32 cm to 34 cm? (Round your answer to two decimal places.)
(b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)
W = (1/2) k x^2
where x is the stretch or compress length
here x = .36 - .28 = .08 meter
so
5 = .5 k (.0064)
k = 1563 N/m so (1/2) k = 781
a)
work to .34 - work to .32
subtract the .28 from both
.06 and .04
= 781 (.06^2-.04^2)
=1.562 J
b) F = k x
15 = 1563 * x
x = .00960 m
= .96 cm
YO YOU'RE WRONG
do Damons thing and divide by 2 at the end.
he is not off on his equations. I followed Damon's instructions to a similar equation and now I understand what I am doing! Thank you for your help!
(a) Well, stretching a spring is a bit like trying to stretch your patience during a math exam. It can be quite a challenge! But don't worry, I'm here to help.
To find how much work is needed to stretch the spring from 32 cm to 34 cm, we first need to determine the initial and final lengths of the spring. The initial length is 32 cm, and the final length is 34 cm. So, we can say that the spring has been stretched by 34 cm - 32 cm = 2 cm.
Now, the work done to stretch the spring is given by the formula W = (1/2)kx^2, where W is the work done, k is the spring constant, and x is the displacement. We know that 5 J of work is needed to stretch the spring by 8 cm (from 28 cm to 36 cm). So, let's plug in the values we know:
5 = (1/2)k(8)^2
Now, we need to find the value of k, the spring constant. Solving the equation, we find that k = 5/(64/2) = 0.15625 J/cm^2.
Finally, let's calculate the work needed to stretch the spring by 2 cm (from 32 cm to 34 cm):
W = (1/2)(0.15625 J/cm^2)(2 cm)^2
W = 0.15625 J/cm^2 * 4 cm^2
W = 0.625 J
So, the work needed to stretch the spring from 32 cm to 34 cm is approximately 0.63 J.
(b) Now, let's calculate how far beyond its natural length a force of 15 N will keep the spring stretched. Remember to wear your math hat tightly because things are about to get stretchy!
We can use Hooke's Law, which states that the force applied to a spring is proportional to the displacement of the spring from its natural length. Mathematically, F = kx, where F is the force, k is the spring constant, and x is the displacement.
We know that the force applied is 15 N, and we need to find the displacement (how far beyond its natural length the spring stretches). So, we rearrange the equation to solve for x:
x = F/k
x = 15 N / 0.15625 J/cm^2
Let's convert the units to cm:
x = 15 N / 0.15625 J/cm^2 * (1 J/1000 mJ) * (100 cm/1 m)
x = 960 cm
So, a force of 15 N will keep the spring stretched approximately 960 cm beyond its natural length.
Remember, these calculations are no joke, but with a bit of math and some humor, we can bounce through them together!
To solve this problem, we can use the formula for potential energy stored in a spring, which is given by:
PE = (1/2)kx^2
where PE is the potential energy, k is the spring constant, and x is the displacement from the natural length of the spring.
To find the spring constant (k), we can use the given information that 5 J of work is needed to stretch the spring from its natural length of 28 cm to a length of 36 cm. The work done on a spring is equal to the change in potential energy:
Work = ΔPE
Since the spring is being stretched, the potential energy increases, so the change in potential energy is positive:
5 J = PE(36 cm) - PE(28 cm)
We need the spring constant (k) to find the potential energy at different displacements. Rearranging the equation, we get:
PE(36 cm) = 5 J + PE(28 cm)
Now, we can calculate the spring constant (k) by using the potential energy at a known displacement and solving for k:
PE = (1/2)kx^2
Rearranging the equation, we get:
k = 2PE / x^2
Using the equation with the given values, we find:
k = 2(5 J) / (36 cm - 28 cm)^2
Now, we have the spring constant (k), and we can use it to answer the questions:
(a) To find the amount of work needed to stretch the spring from 32 cm to 34 cm, we can use the equation for potential energy and substitute the values:
Work = ΔPE = PE(34 cm) - PE(32 cm)
Using the formula for potential energy with the spring constant (k) we calculated:
Work = (1/2)k(34 cm)^2 - (1/2)k(32 cm)^2
Substituting the spring constant (k) value we found earlier:
Work = (1/2)(2(5 J) / (36 cm - 28 cm)^2)(34 cm)^2 - (1/2)(2(5 J) / (36 cm - 28 cm)^2)(32 cm)^2
Simplifying and calculating, we find the answer to be:
Work ≈ 0.63 J (rounded to two decimal places)
(b) To find how far beyond its natural length a force of 15 N keeps the spring stretched, we can rearrange the equation for potential energy:
PE = (1/2)kx^2
We know the force acting on the spring (F) and want to find the displacement (x). Rearranging the equation, we get:
x = sqrt((2PE) / k)
Substituting the spring constant (k) we found earlier:
x = sqrt((2PE) / (2(5 J) / (36 cm - 28 cm)^2))
Simplifying and calculating, we find the answer to be:
x ≈ 0.66 cm (rounded to one decimal place)