In the figure, the 2-kg disk is attached to an elastic cord on a smooth surface, and pushed from rest with the velocity shown. The cord has a “spring” constant of k = 20 N/m, and is initially unstretched. What will its rate of stretching and the speed of the disk be when the cord has been stretched 0.2 m? The image depicts a setup involving an elastic cord and a point mass. ### Description- **Elastic Cord**: Labeled as having a spring constant (\( k_c \)) of 20 N/m.- **Coordinates**: A 3D coordinate system is shown with axes labeled \( x \), \( y \), and \( z \).- **Point of Origin (O)**: The point where the elastic cord is anchored.- **Object at Point D**: A blue circular object is located 0.5 meters away from the origin along the \( x \)-axis.- **Velocity**: The object at point D has an initial velocity (\( v_{D_1} \)) of 1.5 m/s directed along the positive \( y \)-axis.This setup can be analyzed to study the dynamics of the system, particularly how the elastic force affects the motion of the object. The tension in the elastic cord can be calculated using Hooke's Law, provided the extension of the cord from its natural length is known. The velocity vector suggests that the object experiences motion predominantly in the \( y \)-direction at the given moment.

Given: Spring constant of the cord: →k=20 N/m Mass of the disk→m=2 Kg Initially unstretched length of the cord=0.5 m Deformation in the cord after being strectched=0.2 m Initial velocity of the disc→V1=1.5 m/s Because it is given smooth surface so there is no external force hence the total mechanical energy will remains constant for the given system.Initial energy of the system: →ME1=12×m×V12 =12×2×1.52=2.25 J Here the final position of the disc can be shown as below:The final mechanical energy will be equal to initial mechanical energy so : →ME2=ME1→12×k×0.22+12×m×V22=2.25 (where V2 is the final velocity of the disc)→0.5×20×0.04+0.5×2×V22=2.25→V2=1.85=1.36 m/s ...............answer Because angle made by line of cord with the speed of the disc is: →θ=sin-10.50.5+0.2=45.58° Hence the rate of stretching will be: →r=V2×cosθ =1.36×cos45.58° =0.95 m/s .................answer