Math  /  Calculus

Question6. A force of 50.N50 . \mathrm{N} is required to stretch a spring 5.0 cm horizontally from its equitiblum position. How much work is needed to stretch the spring another 2.0 cm from this point?

Studdy Solution

STEP 1

What is this asking? How much energy does it take to stretch a spring that's already stretched out a bit, just a little bit *further*? Watch out! We're dealing with *work* done on a spring, not just force.
Also, the spring is *already* stretched, so we're not starting from zero!

STEP 2

1. Find the spring constant
2. Calculate work for the initial stretch
3. Calculate work for the total stretch
4. Find the difference

STEP 3

Hooke's Law tells us that the force FF needed to stretch a spring is proportional to the distance xx it's stretched: F=kxF = kx, where kk is the **spring constant**.
We're given F=50NF = 50 \, \mathrm{N} and x=5cm=0.05mx = 5 \, \mathrm{cm} = 0.05 \, \mathrm{m} (remember to convert to meters!).

STEP 4

Let's **solve for** kk!
We can divide both sides of F=kxF = kx by xx to get k=Fxk = \frac{F}{x}.
Plugging in our **values**, we get k=50N0.05m=1000N/mk = \frac{50 \, \mathrm{N}}{0.05 \, \mathrm{m}} = 1000 \, \mathrm{N/m}.
That's a stiff spring!

STEP 5

The work done to stretch a spring is given by W=12kx2W = \frac{1}{2}kx^2.
Here, we'll use the **initial stretch** of x=0.05mx = 0.05 \, \mathrm{m} and the kk we just found.

STEP 6

**Plugging in** our values, we have W1=12(1000N/m)(0.05m)2=1.25JW_1 = \frac{1}{2}(1000 \, \mathrm{N/m})(0.05 \, \mathrm{m})^2 = 1.25 \, \mathrm{J}.
So, it took 1.25 J\textbf{1.25 J} of work to stretch the spring those first 5 cm.

STEP 7

Now, the spring is stretched an *additional* 2 cm, for a **total stretch** of 5cm+2cm=7cm=0.07m5 \, \mathrm{cm} + 2 \, \mathrm{cm} = 7 \, \mathrm{cm} = 0.07 \, \mathrm{m}.

STEP 8

Let's **calculate the work** for this total stretch: W2=12(1000N/m)(0.07m)2=2.45JW_2 = \frac{1}{2}(1000 \, \mathrm{N/m})(0.07 \, \mathrm{m})^2 = 2.45 \, \mathrm{J}.

STEP 9

The **extra work** needed to stretch the spring that extra 2 cm is the *difference* between the total work and the initial work: W=W2W1W = W_2 - W_1.

STEP 10

So, W=2.45J1.25J=1.2 JW = 2.45 \, \mathrm{J} - 1.25 \, \mathrm{J} = \textbf{1.2 J}.
Boom!

STEP 11

It takes 1.2 J\textbf{1.2 J} of work to stretch the spring those extra 2 cm.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord