A box is attached to a spring and at rest at the equilibrium position of the spring when a bullet hits the box. The bullet has speed v, prior to hitting the box, it goes completely through the box, coming out the other side with speed v./3. Assume the impact is sufficient to cause to box to start moving. How far is the spring compressed when the box comes to rest (in terms of the given information)? There is friction between the box and the ground. Just give the equation (or equations) needed to solve for the distance. Do not simplify or isolate for the distance, but make sure it is the only unknown. Given information: • Maox (mass of box) • mbullet (mass of bullet) • v, (initial speed of the bullet) • H, (static friction coefficient, box and ground) • H (kinetic friction coefficient, box and ground) k (spring constant) Mbox mpullet

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Spring Potential Energy

The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The potential energy is the measure of the net work done by or on the body. The spring potential energy is the potential energy stored due to the deformation of an elastic spring and this elasticity is due to the stretched spring. The elasticity is the ability of a solid-state matter to come back to its equilibrium position or original position after any kind of external force is acted on it. It is also known as Elastic potential energy. The potential energy here is caused due to the displacement of the spring from its equilibrium position to another position and this potential energy is equal to the net work done due to the stretching of the spring. It also resembles the same mechanics of the oscillation of a simple pendulum, where due to the external force, the object moves from its equilibrium position to its maximum ends and the periodic motion continues till it comes back to the equilibrium position to rest.