Physics - W.E.P. Question with Solution | TestHub

Initially the spring is compressed $x=\frac{5g}{50}=1\mathrm{m}$. After removing 3 kg the 2 kg block's equilibrium is at compression $\frac{2g}{50}=0.4\mathrm{m}$.

(A) Not attached: speed is maximum at equilibrium: $\frac{1}{2}(2)v^2=\frac{1}{2}k(1^2-0.4^2)-2g(0.6)\Rightarrow v^2=9$ - 3 m/s, correct.

(B) It leaves the spring at the natural length with $\frac{1}{2}(2)v^2=\frac{1}{2}k(1{)}^2-2g(1)=5$, rising $h=v^2/2g=0.25\mathrm{m}$ - correct.

(C) Attached: SHM with amplitude 0.6 m about 0.4 m compression, $v_{max}=\omega A=5\times 0.6=3m/s$ - correct. (D) Top of SHM is only 0.2 m above the natural length, not 0.5 m - wrong.

Ans. (A,B,C)