Solution Found!
The spring in figure is compressed by ?x. It launches the
Chapter 10, Problem 40P(choose chapter or problem)
The spring in FIGURE P10.40a is compressed by \(\Delta x\). It launches the block across a frictionless surface with speed \(v_{0}\). The two springs in FIGURE P10.40b are identical to the spring of Figure P10.40a. They are compressed by the same \(\Delta x\) and used to launch the same block. What is the block's speed now?
Equation Transcription:
Text Transcription:
Delta x
v_0
Delta x
Questions & Answers
QUESTION:
The spring in FIGURE P10.40a is compressed by \(\Delta x\). It launches the block across a frictionless surface with speed \(v_{0}\). The two springs in FIGURE P10.40b are identical to the spring of Figure P10.40a. They are compressed by the same \(\Delta x\) and used to launch the same block. What is the block's speed now?
Equation Transcription:
Text Transcription:
Delta x
v_0
Delta x
ANSWER:Step 1 of 2
We have to find the speed of the block launched by the two springs in figure(b) if both the springs are compressed by \(\Delta x\).
The speed of the block launched by the spring in figure \((a)\) is \(\Delta x\).
The initial potential energy of the spring is equal to the final kinetic energy of the block by the law of conservation of energy.
\(\frac{1}{2} k(\Delta x)^{2}=\frac{1}{2} m v_{o}^{2}\)
Where,
\(k=\text { spring constant in } \mathrm{N} / \mathrm{m}\)
\(m=\text { mass of the block in } \mathrm{kg}\)
\(v_{o}=\text { speed of the block in } \mathrm{m} / \mathrm{s}\)