A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
Practice Questions
1 question
Q1
A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
0.5 m
1 m
2 m
0.25 m
Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.
Questions & Step-by-step Solutions
1 item
Q
Q: A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
Solution: Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.
Steps: 7
Step 1: Identify the mass attached to the spring, which is 1 kg.
Step 2: Calculate the force due to gravity acting on the mass using the formula F = mg, where g = 9.8 m/s². So, F = 1 kg * 9.8 m/s² = 9.8 N.
Step 3: Identify the spring constant, which is given as 200 N/m.
Step 4: Use Hooke's law, which states that the force exerted by a spring is F = kx, where k is the spring constant and x is the extension of the spring.
Step 5: Rearrange Hooke's law to find the extension x: x = F/k.
Step 6: Substitute the values into the equation: x = 9.8 N / 200 N/m.
Step 7: Calculate the maximum extension: x = 0.049 m.
