Work Energy And Power Question 307
Question: Two springs of force constant 100 N/m and 150 N/m are in series as shown. The block is pulled by a distance of 2.5 cm to the right from equilibrium position. What is the ratio of work done by the spring at left to the work done by the spring at right.
Options:
A) $ \frac{3}{2} $
B) $ \frac{2}{3} $
C) 0.2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ k_1x_1=k_2x_2 $
$ 100x_1=150x_2 $
and $ x_1+x_2=2.5\Rightarrow x_2=1cmx_1=1.5cm $
$ W.D.=-\frac{1}{2}( x_f^{2}-x_1^{2} ) $ ;
Ratio $ \frac{-\frac{1}{2}100\times {{(1.5)}^{2}}}{-\frac{1}{2}150\times {{(1)}^{2}}}=\frac{3}{2} $
BETA
