During the physics lab, a student finds that a certain spring pulls with a force of 2.4 newtons when it is stretched out by 0.6 cm. The student divides 0.6/2.4 to obtain the number 0.25. Which of the following is correct?
a. The spring will generate 0.25 newtons of force for each centimeter that it is stretched.
b. The number 0.25 has no physical interpretation; essentially meaningless.
c. The spring must be stretched by 0.25 centimeters to generate 1 newton of force.
The correct answer and explanation is:
The correct answer is b. The number 0.25 has no physical interpretation; essentially meaningless.
Here’s why:
The student attempts to calculate a relationship between force and displacement by dividing the amount of stretch (0.6 cm) by the force (2.4 newtons), obtaining the value 0.25. However, this value does not have any valid physical meaning in the context of Hooke’s law or the spring constant.
Hooke’s law states that the force exerted by a spring is directly proportional to its displacement, given by the equation:
F=kxF = kx
Where:
- FF is the force exerted by the spring (in newtons),
- kk is the spring constant (in newtons per meter),
- xx is the displacement or stretch of the spring (in meters).
From the information provided:
- The force is 2.4 newtons.
- The displacement is 0.6 cm, which needs to be converted into meters (0.6 cm = 0.006 m).
To calculate the spring constant kk, we can rearrange Hooke’s law to solve for kk: k=Fx=2.40.006=400 N/mk = \frac{F}{x} = \frac{2.4}{0.006} = 400 \, \text{N/m}
Now, the student’s approach of dividing 0.6 by 2.4 gives a value of 0.25, but this does not provide any useful information. It does not relate to the spring constant, nor does it give a meaningful interpretation of how the spring behaves. Dividing displacement by force like this is not how spring behavior is characterized in physics.
For a valid interpretation, the correct method would involve finding the spring constant first, which gives the relationship between force and displacement for that particular spring.