Work Energy And Power Question 280
A block lying on a smooth surface with spring connected to it is pulled by an external force as shown. Initially the velocity of ends A and B of the spring are 4 m/s and 2 m/s respectively. If the energy of the spring is increasing at the rate of 20 J/sec, then the stretch in the spring is 0.2 m
Options:
A) 1.0 cm
B) 2.0 cm
C) 10 cm
D) 2.0 cm
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ {x _{A}} and x _{B} $ be the position of ends A and B at time (from the block, then stretched length of the spring will be $ l_2=x _{A}-x _{B} $ and so the stretch $ \Delta \ell ={\ell_2}-{\ell_1}=(x _{A}-x _{B})-{\ell_1}({\ell_1} $ natural length of the spring) So, $ U=\frac{1}{2}k\Delta {{\ell }^{2}}=\frac{1}{2}k{{[(x _{A}-x _{B})-{\ell_1}]}^{2}} $
$ P=\frac{dU}{dt}=\frac{1}{2}k\cdot 2(x _{A}-x _{B}-{\ell_1})( \frac{dx _{A}}{dt}-\frac{dx _{B}}{dt} ) $
$ P=F( v _{A}-v _{B} )F=\frac{d x _{A}}{v _{A}-v _{B}} $
$ \Delta \ell =\frac{F}{k}=\frac{P}{(v _{A}-v _{B})}=\frac{20}{4-2\times 10^{-2}} $
$ \Delta \ell =0.1m=10cm $
BETA
