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Hooke's Law, Work and Elastic Potential Energy

The more you stretch (or compress) a spring, the force required to stretch (or compress) it further gets larger. For small deformations of a spring, the force needed to create that deformation is proportional to the deformation: F = kx. This is Hooke's Law, where k is the spring constant and x is the deformation of the spring (the distance the spring has been squeezed or stretched). Elastic potential energy is the work associated with deforming an object, such as a spring. The elastic potential energy of a spring is (.5)kx2.

Elastic Potential Energy Problems

(See bottom of page for answers.)

1. When a 13.2-kg mass is placed on top of a vertical spring, the spring compresses 5.93 cm. Find the force constant of the spring.

2. If a spring has a spring constant of 400 N/m, how much work is required to compress the spring 25.0 cm from its undisturbed position?

3. A compressed spring that obeys Hooke's law has a potential energy of 18 J. If the spring constant of the spring is 400 N/m, find the distance by which the spring is compressed.

4. An object is attached to the lower end of a 100-coil spring that is hanging from the ceiling. The string stretches by 0.165 m. The spring is then cut into two identical springs of 50 coils each. Each spring is attached between the ceiling and the object. By how much does each spring stretch?

5. A vertical spring stretches 10 cm under a load of 200 g.
a. Determine the spring constant.
b. How much work is required to stretch the first 5 cm.
c. How much work is required to stretch the last 5 cm.

6. A mass sitting on a horizontal frictionless surface is attached to one end of spring; the other end of the spring is fixed to a wall. To compress the spring by 0.12 m requires 3.0 J of work. If the mass is released from rest with the spring compressed, it experiences a maximum acceleration of 15 m/s/s. Find the value of the spring constant.

7. To compress spring 1 by 0.20 m takes 150 J of work. Stretching spring 2 by 0.30 m requires 210 J of work. Which spring is stiffer?

Answers to Elastic Potential Energy Problems

Selected solutions are printed below.
For solutions to all the problems on this page click here.

1. From Hooke's Law: F = kx
The force on the spring is the weight of the object, 13.2*9.8 = 129 N
x = 5.93 cm =0.0593 m
129 = k(0.0593)
k = 2181 N/m

For solutions to all the problems on this page click here.

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