Question

An ideal spring of unstretched length 0.2m is placed horizontally on a frictionless table as shown below. One end of the spring is fixed and the other end is attached to a block of mass M=8.0kg. The 8kg block is also attached to a massless string that passes over a small frictionless pulley. A block of mass m=4kg hangs from the other end of the string. When the spring and blocks system is in equilibirium, the length of the spring is 0.25m 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 800 N
• B. Force constant of the spring is 800 N/m
• C. Time taken by the 4 kg block to hit the floor after $t_0$ is 1 sec
• D. Maximum speed achieved by 8.0 kg block is 5m/s

Answer 1

Answer: (B), (C)

The correct options are
B Force constant of the spring is 800 N/m
C Time taken by the 4 kg block to hit the floor after $t_0$ is 1 sec

T=4g=40N $⟶option\left(A\right)\times$
$\therefore Kx=40N\therefore K(0.25-0.2)=40\therefore K\left(0.05\right)=40\therefore K=800N/m⟶option\left(B\right)\surd S=ut+\frac{1}{2}a{t}^2\therefore 5=0+\frac{1}{2}\left(10\right)t^2\therefore 10=10({t)}^2\therefore t=1sec⟶option(C)\surd ForMaximumspeed;\frac{1}{2}K{x}^2=\frac{1}{2}m{v}^2\therefore K{x}^2=m{v}^2\therefore (800\left)\right(0.05{)}^2=8(v{)}^2\therefore v=0.5m/s⟶option(D)\times$