The block shown weighs 15 lb. The spring has a spring constant k of 10 lb/in. If the force F is 50 lb, how much longer is the spring than its unstretched length? A. 4.5 in B. 2.8 in C. 5.2 in D. 1.0 in E. None of the above F 57.3°
The Correct Answer and Explanation is:
To solve this problem, we need to find how much longer the spring is stretched from its unstretched length when a force of 50 lb is applied. The force exerted by the spring is given by Hooke’s Law: F=k⋅ΔxF = k \cdot \Delta xF=k⋅Δx
Where:
- FFF is the force applied to the spring (in pounds).
- kkk is the spring constant (in lb/in).
- Δx\Delta xΔx is the displacement or extension of the spring from its unstretched length (in inches).
Given:
- F=50F = 50F=50 lb
- k=10k = 10k=10 lb/in
We can rearrange Hooke’s Law to solve for the displacement: Δx=Fk\Delta x = \frac{F}{k}Δx=kF
Now, substituting the given values: Δx=50 lb10 lb/in=5 in\Delta x = \frac{50 \, \text{lb}}{10 \, \text{lb/in}} = 5 \, \text{in}Δx=10lb/in50lb=5in
Thus, the spring is stretched by 5 inches longer than its unstretched length.
The correct answer is E. None of the above, because none of the choices (A, B, C, D) match 5 inches.
Explanation:
This problem involves applying Hooke’s Law, which describes how the displacement of a spring is related to the force applied to it. The spring constant kkk is a measure of the stiffness of the spring, with larger values of kkk meaning the spring is stiffer and harder to stretch. By dividing the applied force FFF by the spring constant kkk, we determine the extension Δx\Delta xΔx, which gives us how much longer the spring becomes under the applied force.
In this case, the spring stretches by 5 inches when a 50 lb force is applied.
