A body of mass 0.025 kg attached to a spring is displaced through 0.1 m to the right of the equilibrium position. If the spring constant is 0.4 N/m and its velocity at the end of displacement is 0.4 m/s, then its total energy will be:
Added by Lisa D.
Step 1
025 \) kg - Displacement from equilibrium, \( x = 0.1 \) m - Spring constant, \( k = 0.4 \) N/m - Velocity at the end of displacement, \( v = 0.4 \) m/s Show more…
Show all steps
Close
Your feedback will help us improve your experience
Adi S and 58 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)
Adi S.
A spring of spring constant k = 8.25 N/m is displaced from equilibrium by a distance of 0.150 m. What is the stored energy in the form of spring potential energy? Show your work.
Prabhu R.
Shalin L.
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD