Work Energy And Power Question 227
Question: A block of mass $ m=0.1 $ kg is connected to a spring of unknown spring constant k. It is compressed to a distance $ ( \frac{x}{2} ) $ from its equilibrium position and released from rest. After approaching half the distance from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity $ 3m{s^{-1}} $ . The total initial energy of the spring is
Options:
A) $ 0.3J $
B) $ 0.6J $
C) $ 0.8J $
D) $ 1.5J $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Applying momentum conservation $ {m_1}u_1 + m_2u_2 = {m_1}v_1 + m_2v_2 $
$ 0.1u + m(0) = 0.1(0) + m(3) $
$ 0.1\mu=3\text{m}$
$ \frac{1}{2}(0.1)u^{2}=\frac{1}{2}m{(3)}^{2} $ Solving we get, $ u=3\sqrt{m} $
$ \frac{1}{2}kx^{2}=\frac{1}{2}K{{( \frac{x}{2} )}^{2}}+\frac{1}{2}( 0.1 )v^{2} $
$ \Rightarrow \frac{3}{4}kx^{2}=0.9\Rightarrow \frac{3}{4}kx^{2}=0.9 $
$ \therefore \frac{1}{2}Kx^{2}= 0.6 J $ (total initial energy of the spring)
BETA
