The energy stored in a stretch spring depends on the distance x that it has been stretch according to the equation E=(1/2)kx^2 where k is a constant that depends on how stiff the spring is

Question

The energy stored in a stretch spring depends on the distance x that it has been stretch according to the equation E=(1/2)kx^2 where k is a constant that depends on how stiff the spring is. Suppose that for a certain spring, k=0.20. This spring is being stretched at a rate of 1.5 cm/sec. How quickly is the energy stored in the spring increasing at the moment that the spring has been stretched x = 10 cm?

Answer 1

The bottom line for this problem is: 'what is the rate of change of the stored energy in the spring for a given distance from equilibrium (not stretched)?' So you are looking for the slope in the function E(t) for a given tn that is governed by the rate of the deformation dx/dt. You have a constant rate of deformation r= dx/dt so on integrating x (deformation) distance is related to a deformation time as t = x/ r. Since E(t) and x(t) are continuous differentiable functions. We can differentiate dE(t)/dt as dE(t)/dx x dx/dt substituting E(t) for E(x) giving dE/dt = dE(x)/dt = d E(x)/dx (dx/dt). This is the tangent to the E(t) curve for the time point related to x(t) equal to10 cm.

So we take the derivative of E(x) relative to x(t) and multiply by dx(t)/dt being a constant rate r substituting the current deformation of 10 cm. Plotting dE/dt is of the same form as dE/dx save the constant rate in the length deformation of the spring.

I believe that this problem was to illustrate for a generalized coordinate qs,

df(q)/dqj=df(q)/dqi (dqi/dqj) for q in this case as x.t .

This problem can give a bit more insight into the chain rule:

It could help to plot some functions to see this....

Perhaps g(u) with u = x ^2 and g (u) = u^2 and find dg/du vs dg/dx

Perhaps another way to state the chain rule is that rate in change of the function relative to a coordinate is equivalent to the change relative to another coordinate multiplied by the change of that coordinate relative to the first coordinate. If dx/dt is 1 these rates are the same. If dx/dt is c these rates just a multiple of c for all x and t. If dx/dt = f(x) Then dE/dt = dE/dx f(x) for a given x and the plot of this function would not be of the same form as dE/dx,

Answer 2

1. Energy Stored in a Stretched Spring

The energy (E) stored in a stretched spring is given by the formula:

E = (1/2) k x²

where:

- E is the energy stored in the spring (in Joules, J),

- k is the spring constant (in N/m),

- x is the displacement from the equilibrium position (in meters, m).

2. Given Information

- Spring constant, k = 0.20 N/m

- Stretching rate, dx/dt = 1.5 cm/s = 0.015 m/s (converted to meters per second)

- Displacement, x = 10 cm = 0.1 m (converted to meters)

3. Calculating Instantaneous Energy

The energy stored in the spring at a stretch of 10 cm can be calculated using the formula:

E = (1/2) * k * x²

Substituting the given values:

E = (1/2) * 0.20 N/m * (0.1 m)²

E = (1/2) * 0.20 N/m * 0.01 m²

E = (1/2) * 0.002 J

E = 0.001 J

Thus, the energy stored in the spring at 10 cm stretch is 0.001 Joules.

4. Rate of Change of Energy Stored in the Spring

To find how quickly the energy stored in the spring is increasing, we need to find the time derivative of the energy (dE/dt). Using calculus, the time derivative of the energy is given by:

dE/dt = d/dt [(1/2) k x²]

Applying the chain rule, we get:

dE/dt = k * x * (dx/dt)

where dx/dt is the rate at which the spring is being stretched.

5. Calculation

Substituting the values:

dE/dt = 0.20 N/m * 0.1 m * 0.015 m/s

dE/dt = 0.20 * 0.0015 J/s

dE/dt = 0.0003 J/s

Therefore, the rate of change of energy stored in the spring is 0.0003 Joules per second.

6. Interpretation

The positive rate of change of energy stored in the spring (0.0003 J/s) indicates that the energy is increasing at this rate when the spring is stretched to 10 cm and is being stretched at a rate of 1.5 cm/sec.