Work Energy And Power Question 214
A spring of spring constant $ 5\times 10^3 N/m $ is stretched initially by 5 cm from the upstretched position. Then the work required to stretch it further by another 5 cm is
Options:
A) 18.75 J
B) 25.00 J
C) 6.25 J
D) 12.50 J
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ {W_1}=\frac{1}{2}\times 5\times 10^{3}{{( 0.05 )}^{2}} $
$ \Rightarrow W_2=\frac{1}{2}\times 5\times 10^{3}{{( 0.10 )}^{2}} $
$ \therefore \Delta W=\frac{1}{2}\times 5\times 10^{3}\times 0.15\times 0.05 =18.75J. $
BETA
