Copyright

A horizontal spring is compressed a distance 8cm from its equilibrium position and released...

Question:

A horizontal spring is compressed a distance 8cm from its equilibrium position and released against a ball of mass 50 g which is propelled away from the spring at 5 ms-1.

A) Calculate the spring constant.

B) Calculate the work in compressing the spring.

Restoring force:

The restoring force experienced by the spring is given by,

{eq}F = -kx {/eq}

Where,

F = force

k = spring constant

x = compression

The work done on the spring is,

{eq}W =\dfrac{1}{2} kx^{2} {/eq}

Where,

W = work done

Answer and Explanation: 1

Given:

Mass of ball, m = 50 g = 0.05 kg

Speed of the spring, v = 5 m/s

Compression, x = 8 cm = 0.08 m

Calculation:

A)

The restoring force on the spring is given by,

{eq}F = -kx {/eq}

The negative sign of the force is due to opposite movement of the spring.

The restoring force is balanced by the weight of the ball. Thus,

{eq}mg = kx {/eq}

{eq}k = \dfrac{mg}{x}= \dfrac{0.05 \times 9.8}{0.08}= 6.125 N/m {/eq}

The spring constant is 6.125 N/m.

B)

The work in compressing the spring is given by,

{eq}W = \dfrac{1}{2} kx^{2} = \dfrac{1}{2} 6.125 \times 0.08^{2} = 0.0196 J {/eq}

The work done in compressing spring is 0.0196 J.


Learn more about this topic:

Loading...
Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
202K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.


Related to this Question

Explore our homework questions and answers library