If it takes 4.12 J of work to stretch a Hooke's law spring 12.1 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm.

The stress-strain curve for a Hooke's law spring is a triangle. The area is the work done. Therefore the work done is proportional to the square of the displacement from unstressed state.

4.12 J : 12.1²
X : (12.1+10)²
Can you take it from here?

No

To determine the extra work required to stretch the Hooke's law spring an additional 10.0 cm, we can first calculate the spring constant using the initial information given.

Hooke's Law states that the force exerted on a spring is proportional to the displacement from its equilibrium position. Mathematically, it is expressed as:

F = -kx

Where:
F is the force exerted on the spring
k is the spring constant
x is the displacement from the spring's unstressed length

From the given information, we know that the work required to stretch the spring 12.1 cm is 4.12 J. We can calculate the spring constant using this information.

The formula for the work done on a spring is given by:

W = (1/2)kx^2

Rearranging the equation, we can solve for the spring constant:

k = (2W) / x^2

Substituting the given values:

k = (2 * 4.12 J) / (0.121 m)^2

Now that we have the spring constant, we can calculate the additional work required to stretch it by 10.0 cm.

First, we need to convert the displacement from cm to m:

Displacement = 10.0 cm = 0.1 m

Using the formula for work done on a spring, we have:

W_extra = (1/2)k * (displacement^2)

W_extra = (1/2) * k * (0.1 m)^2

Plugging in the value of k we calculated earlier:

W_extra = (1/2) * (2.176 m^(-1)) * (0.1 m)^2

Finally, compute the additional work:

W_extra = (1/2) * (2.176 m^(-1)) * (0.01 m)

W_extra = 0.0218 J

Therefore, the extra work required to stretch the Hooke's Law spring an additional 10.0 cm is approximately 0.0218 J.

To determine the extra work required to stretch the Hooke's law spring an additional 10.0 cm, we first need to understand Hooke's law.

Hooke's law states that the force required to stretch or compress a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:

F = k * x

Where:
F is the force applied to the spring,
k is the spring constant, and
x is the displacement of the spring.

In this case, we know that it takes 4.12 J of work to stretch the spring by 12.1 cm. Work can be calculated by the equation:

Work = Force * Distance

We can rearrange the Hooke's law equation to solve for the force:

F = k * x

Since the work and displacement are given, we can find the force applied:

4.12 J = F * 0.121 m

Solving for F:

F = 4.12 J / 0.121 m
F ≈ 34.05 N

Now that we know the force, we can find the spring constant (k). Rearranging the Hooke's law equation:

F = k * x

k = F / x
k = 34.05 N / 0.121 m
k ≈ 281.07 N/m

With the spring constant determined, we can now calculate the extra work required to stretch the spring an additional 10.0 cm. The displacement (x) is 10.0 cm, which is 0.1 m.

Using the work equation:

Work = Force * Distance

We can calculate the extra work:

Work = (k * x) * Distance
Work = (281.07 N/m * 0.1 m) * 0.1 m
Work ≈ 2.81 J

Therefore, the extra work required to stretch the Hooke's law spring an additional 10.0 cm is approximately 2.81 J.