Energy

Elastic PE and the Spring Constant

Hooke's Law, the Spring Constant k, and Elastic Potential Energy

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Do Now warm-up prompt

Spring Constant (k) - a measure of a spring's stiffness; how much force is needed to stretch or compress it a given distance

What is the shape of a plot that shows a direct square relationship?

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Fill in the Blank Notes

Review - Spring Constant

Spring constant review diagram

PE and k

e) PEs in terms of k — Ideal Spring

1. PEs = Fx

Average Force: F = F/2

2. PE = (F/2)x

substitute F = kx (on reference)

3. PEs = (1/2)(kx)x

(on reference)

PEs = 1/2(kx2)

Ex 1

Given

A force of 12. N stretches a spring and makes it .15 m longer.
Find: a) the spring constant (k) of this spring.

Spring stretching under an applied force
↓

k = F/x
= 12. N / .15 m

k = 80. N/m

Spring diagram Spring diagram Spring diagram

Ex 1 (cont.)

Given

b) What is the PE of this spring?

↓

PEs = (1/2)kx2

= 1/2(80. N/m)(.15 m2)

Be sure to use parenthesis when calculating!!

PEs = .90 Joules

Bungee jumper attached to a stretched spring

Ex 2

Given

If 15. joules of energy are stored in the stretched spring below, what is the value of the spring constant?

Spring stretched by 0.50 m storing 15 joules of elastic PE

(on reference)

PEs = 1/2(kx2)

↓

15. J = (1/2)k(.50 m)2

(1/2 in decimal form is .50)

15. J = .50k(.25)

15 = .125k

Divide both sides by .125

k = 120 N/m

What's Wrong With This Picture?

PEs = (1/2)kx2   (x = elongation)

Incorrect PE versus x graph drawn as a straight line

PEs = 1/2(kx2) — PE, x direct square

Plot should be curved

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Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. A force of 12 N stretches a spring 0.15 m. What is the spring constant?

2. A spring with k = 80. N/m is stretched 0.15 m. How much elastic PE does it store?

3. A spring stretched 0.50 m stores 15 J of energy. What is its spring constant?

4. If the stretch of an ideal spring is doubled, its elastic potential energy becomes

5. What is the shape of a graph of elastic PE (y-axis) versus stretch x (x-axis) for an ideal spring?

NYSSLS Regents Practice · Energy · Energy Transformations

A Toy Dart Launcher

A toy dart launcher uses a spring with a spring constant of 400. N/m. The spring is compressed 0.050 m and then released, launching a 0.010-kg foam dart. Friction is negligible.

How much elastic potential energy is stored in the compressed spring?

▶ Want a challenge? Open 6 more Regents-style questions  (harder · mixed formats)

Challenge score: 0 / 6

Substitute values

C1. All 0.50 J of the stored elastic potential energy becomes kinetic energy of the dart. Substitute numerical values into a mathematical representation that can be used to determine the dart’s launch speed. Then calculate the speed.

v = √(2( J)( kg)) = m/s
Multiple choice

C2. The spring is compressed twice as far (0.10 m). How does the stored energy change?

Draw the vector · Determine the magnitude

C3. The launcher is held level and points to the right. At the instant the compressed spring is released, choose the direction of the force the spring exerts on the dart, and determine its magnitude.

dart
●
Fs = N
Support or Refute · Justify

C4. A student makes the following claim.

Claim: A heavier 0.020-kg dart launched from the same 0.050-m compression would leave the launcher at half the speed, 5.0 m/s.

Place a check mark in either the Support or Refute box. Justify your response using a mathematical relationship.

Multiple choice

C5. On a graph of spring force versus compression, what does the area under the line from 0 to 0.050 m represent?

Constructed response

C6. To make darts fly farther, the company considers replacing the spring with one that has k = 800. N/m, compressed the same 0.050 m. Identify one benefit and one drawback of this design change.

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