A spring stretches 5 cm when a load of 10 N is applied. What is the spring constant?
Practice Questions
1 question
Q1
A spring stretches 5 cm when a load of 10 N is applied. What is the spring constant?
200 N/m
100 N/m
50 N/m
25 N/m
Using Hooke's law, k = F/x = 10 N / 0.05 m = 200 N/m.
Questions & Step-by-step Solutions
1 item
Q
Q: A spring stretches 5 cm when a load of 10 N is applied. What is the spring constant?
Solution: Using Hooke's law, k = F/x = 10 N / 0.05 m = 200 N/m.
Steps: 7
Step 1: Understand that Hooke's law states that the force (F) applied to a spring is directly proportional to the extension (x) of the spring. The formula is F = k * x, where k is the spring constant.
Step 2: Identify the values given in the problem. The force (F) is 10 N and the extension (x) is 5 cm.
Step 3: Convert the extension from centimeters to meters because the standard unit for length in physics is meters. 5 cm is equal to 0.05 m.
Step 4: Rearrange Hooke's law to find the spring constant (k). The formula becomes k = F / x.
Step 5: Substitute the values into the formula. k = 10 N / 0.05 m.
Step 6: Calculate the value of k. 10 N divided by 0.05 m equals 200 N/m.
Step 7: Conclude that the spring constant is 200 N/m.
