The 5 -lb cylinder is falling from A with a speed vA=10 ft / s onto the platform. Determine the

The 5 -lb cylinder is falling from A with a speed vA=10 ft /...

The 5 -lb cylinder is falling from $A$ with a speed $v_{A}=10 \mathrm{ft} / \mathrm{s}$ onto the platform. Determine the maximum displacement of the platform, caused by the collision. The spring has an unstretched length of $1.75 \mathrm{ft}$ and is originally kept in compression by the 1 -ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision.

Key Concepts

Conservation of Mechanical Energy

This principle states that in a closed system without non?conservative forces (like friction or air resistance), the sum of kinetic energy, gravitational potential energy, and elastic potential energy remains constant. In collision problems where energy losses are neglected, the energy of a moving object can be completely transferred into the potential energy stored in a deformed spring, allowing one to relate the speed of the falling object to the spring’s displacement.

Elastic Potential Energy (Hooke's Law)

Hooke's Law describes how a spring exerts a force proportional to its displacement from its equilibrium position, and the elastic potential energy stored in a spring is given by one half times the spring constant times the square of the displacement. This concept is crucial when analyzing systems where mechanical energy is temporarily stored in deformable elements like springs during impacts.

Energy Transfer in Impacts

When an object collides with a platform supported by a spring, its kinetic energy is converted into elastic energy as the spring compresses. Understanding the transfer and transformation of energy in such collisions allows for the analysis of maximum displacements and subsequent motion of the system, provided that energy losses are negligible.

Transcript

00:01 In this problem, we have a block that weighs 5 pounds and travels at 10 feet per second that strikes a platform that has a spring underneath.

00:09 And we want to find the distance that this spring compresses after the collision with the block.

00:15 Now, its weight of 5 pounds, we can convert to mass by divided by 32 .2 feet per second squared.

00:22 And we're also given the stiffness of the spring, which is k is equal to 400 pounds per feet.

00:28 Now because we're dealing with the velocity, this block has kinetic energy and it also has a weight.

00:35 So there's a force being produced as it travels before it strikes the spring.

00:41 And the spring itself has a force, so it will produce a work in the opposite direction of this block when it collides.

00:50 So because we're dealing with work and we're dealing with energy, we can use the work energy principle, which is the initial kinetic energy plus the work being done by the forces is equal to the final kinetic energy.

01:08 But because the block stops moving after the collision, its final kinetic energy will be zero because the velocity will be zero.

01:18 So we're left with the initial kinetic energy, one -half times the mass, times the velocity, we'll call v1 squared, plus.

01:29 Plus the work being done by the forces equals zero.

01:35 We'll call this equation one.

01:39 So now let's look at the work being done by the forces.

01:45 So the forces involved are the weight due to the block and the force due to the spring.

01:53 So let's look first at the block.

01:56 So work is force times distance.

01:58 So we have the weight of this block times the distance.

02:02 So we know it travels three feet, which we'll call s -a, plus the distance this spring compresses after the collision.

02:13 Okay, that's for the block.

02:14 Now let's look at the spring.

02:16 And again, the spring will create a work in the opposite direction of this block...

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