Question

The system is released from rest with the spring initially stretched $75$ $mm$. Calculate the velocity $v$ of the block after it has dropped $12$ $mm$. The spring has a stiffness of $1050$ $N/m$. Neglect the mass of the small pulley.

• A. $v=0.371m{s}^{-1}$
• B. $v=0.471m{s}^{-1}$
• C. $v=0.521m{s}^{-1}$
• D. $v=0.571m{s}^{-1}$

Answer 1

The correct option is A $v=0.371m{s}^{-1}$

$Iftheblockgoesdownbyx,thenthespringisfurtherelongatedby2x.$
$Applyingconservationofenergy:$
$-\frac{1}{2}k{x_1}^2=-mgx+\frac{1}{2}m{v}^2-\frac{1}{2}k{(x_1+2x)}^2$
$x_1=0.075m,x=0.012m,m=45kg,k=1050N/msolvingwegetv=0.371m/s$