An M = 1.35 kg block is placed on top of a vertical spring, whose force constant is k = 12,400 N/m.  The spring is compressed by d = 0.0230 m and is initially at rest.  When the block is released, it accelerates upward. (Insert Image) (1)  Find the speed of the block when the spring has returned to the spring's equilibrium position. (2)  Upon reaching the equilibrium position, the block leaves the spring and continues to rise upward. Ignore any oscillation of the spring.  Find the maximum height reached by the block relative to the equilibrium position of the spring.

Q: A 0.150 kg block is placed on a light vertical spring (k = 5.00 ✕ 103 N/m) and pushed downward,…

A: Given: A 0.150 kg block is placed on a light vertical spring (k = 5.00 ✕ 103 N/m) and pushed…

Q: A 2.4 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it…

A: Using energy conservation  KE=PE stored in spring

Q: A 3.1 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it…

A: we have to given  mass of block = 3.1 kg   initial speed = 1.8 m / s  because surface is friction…

Q: The spring is compressed a mass (m 450 g) a distance of Ax = 9.0 cm. When the mass is released from…

A:

Q: A 1.8 kg block is pushed up against a spring whose spring constant is 650 N/m, compressing it 5.1…

A: Given that, the mass of the block is m=1.8kg spring constant is k=650N/m maximum compression…

Q: A block of mass 2.80 kg is placed against a horizontal spring of constant k = 805 N/m and pushed so…

A: a)The required elastic potential is,

Q: A block of mass 3.40 kg is placed against a horizontal spring of constant k = 855 N/m and pushed so…

A: mass, m = 3.4 kg  spring constant, k = 855 N/m  compression, x = 0.065 m

Q: 64 ing on a horizontal, frictionless table. The block is pushed 5- o the spring, compressing it by…

A:

Q: A horizontal spring on a frictionless surface with a spring constant of 40 N/m is compressed 0.3 m…

A: Given x=0.3m K=40N/m Mass= 2kg Stored energy in spring, U=12kx2

Q: Spiderman uses his spider webs to save a runaway train moving about 17.0 m/s. His web stretches a…

A: Initial velocity of train, u=17 m/s Distance covered, s=500 m Mass of train, m=150000 kg Let spring…

Q: A 306 g mass slides down a curved incline then collides with a spring. The spring constant of the…

A: Mass (m) = 306 g   = 0.306 kg   Spring constant (k) = 156 N/m  Compression of spring (d) = 0.63 cm…

Q: The 5.0-kg object is released from rest at a height of 3.00 m on a curved frictionless ramp. At the…

A:

Q: A block with mass of 0.85 kg is released at the top of a frictionless ramp, 1.2 m above the floor.…

A: mass = 0.85 Kg H = 1.2 m Initial potential energy = mgH   = 0.85 * 9.8 * 1.2  = 9.996 Joules…

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.