Question

An ideal spring of unstreched length of 0.2m is placed horizontally on a friction less table as shown below.One end of the spring is fixed ad the other end is attached to a block of mass M= 8 kg. The 8kg vlock is also attached to a massless string that passes overa small friction less pulley. A block of mass= 4kg hangs from other end of the string.When the spring and block system in equilbrium,the length of the spring is.25m and the 4kg block is 5m above the floor.At $t=t_0$,string is cut at point P, then
(A)Tension in the string before $t_0$ is 800N.
(B)Force constsnt of the spring is 800N/m.
(C)Time taken by the 4kg block to hit the floor after $t_0$ is 1 sec.
(D)Maximum speed achieved by 8kg block is 5m/s.

Answer 1

$T=40N=F_{s}kr=F_{s}kx=40⟹k=\frac{40}{(0.25-0.2)}=\frac{40}{0.05}=800N/ms=ut+1/2a{t}^{2}5=1/2\times 10t^2=t=1sec△\left(\frac{1}{2}k{x}^2\right)=△\left(\frac{1}{2}m{v}^2\right)=800\times {0.25}^2-800\times {.2}^2=8v^{2}v=1.5m/s$