Question

Use calculus to solve a physics question involving work. Suppose 2J of work needed to stretch a spring from its natural length of 32cm to a length of 42cm. (a) How much work needed to stretch the spring from 34cm to 38cm? (b) How far beyond its natural length will a force of 35 N keep the spring stretched?

Answer 1

Have you thought about which equations (from Physics) would be relevant here? For example, the equation for Force (F) is F(x) = kx, where k is the spring constant and x is the net displacement (new position - old position).

Answer 2

Work is the integral of [f(x) times dx].

Answer 3

To find K:
\[42cm-32cm = 10cm = 0.1m\]
\[[\frac{ 1 }{ 2 }kx^2]=2\]
\[k(0.1)^2=4 = \frac{ 4 }{ 0.01 } = 400 = k\]
\[38cm-34cm = 4cm = 0.04m \]
\[38cm-32cm = 6cm = 0.06m\]
\[W=200[x^2]_{0.04}^{0.06}=200[(0.06)^2-(0.04)^2] =0.4\]
My attempt but the answer is wrong.

Answer 4

Everything looks great here, you found your 'k' which I also have to be 400

However, I believe your bounds are messed up a bit

You want your upper bound to be \(\large 38 - 32\) and lower bound to be \(\large 34 - 32\) You always refer back to the natural length

So I have \(\large \int_{0.02}^{0.06} 400xdx = 0.64J\)

Answer 5

\[W=\int\limits_{0.04}^{0.06}400xdx = \int\limits_{0.04}^{0.06}\frac{ 400 }{ 2 }x^2=\int\limits_{0.04}^{0.06}200x\]
So, I simplified that last part wrong?

Answer 6

\[[200x]_{0.04}^{0.06}\]

Answer 7

opps,\[ 200x^2\]**

Answer 8

Nope your bounds are the problem

You had the bounds as LOWER: \(\large 38cm - 34cm\) and UPPER: \(\large 38 - 32cm\)
Your upper bound is FINE, but the lower should be \(\large 38cm - 32cm\) because you always refer back to the natural length

Look at your example for reference as well...Look at the bounds they got there

35 - (natural Length = 30)
and
40 - (natural length = 30)

Answer 9

Okay, thanks for the clarification on part(a). Completely got it now. For part(b), I would do:

400x = 35 and solve for x?

Answer 10

Absolutely!

Answer 11

okay, I got 11.43 -- How would I express it as cm?

Answer 12

Check that again...remember you're solving for 'x'

\[\large 400x = 35\]
\[\large x = \frac{35}{400}\]

Answer 13

Ahh, yes, which is 0.0875 = 0.09 (round to two decimals).

Answer 14

Yeah if they tell you to round, that's correct!

Answer 15

Now, how would I express the answer in cm?

Answer 16

Well remember \(\large 1m = 100cm\)

so \(\large 0.0875m = ?cm\)

Answer 17

Okay, I got part(b) now. Thanks for the help! :)

Answer 18

Not a problem :)