**Problem Statement:**A box with a mass of 18 kg slides down an inclined plane that is 2.0 m high and 3.0 m long along the inclined plane. Due to friction, the box reaches a velocity of 4.2 m/s at the bottom of the inclined plane. Beyond the inclined plane lies a spring with a spring constant of 400 N/m. The spring is fixed at its right end. The level ground between the incline and the spring has no friction.The box compresses the spring and then gets pushed back towards the incline by the spring. How far along the inclined plane, from the bottom, will the box temporarily stop on the inclined plane?**Relevant Data:**1. Mass of the box (m): 18 kg2. Height of the inclined plane (h): 2.0 m3. Length of the inclined plane (L): 3.0 m4. Velocity of the box at the bottom (v): 4.2 m/s5. Spring constant (k): 400 N/m**Answer:**To find the point at which the box will temporarily stop on its way back up the inclined plane, we need to solve for the distance the box travels along the plane after being pushed by the spring. This involves using principles from mechanics such as energy conservation and kinematics, taking into account the forces acting on the box due to gravity, friction, and the restoring force of the spring.1. Calculate the work done by the spring on the box.2. Apply energy conservation to relate this work to the gain in potential energy and loss in kinetic energy as the box moves up the incline.By solving the equations, we can determine the distance along the inclined plane where the box will momentarily come to a stop before potentially sliding back down or remaining at rest depending on the friction. The key equations involve the conservation of mechanical energy, the work-energy theorem, and the understanding of spring dynamics.

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A box with 18 kg of mass slides down an inclined plane that is 2.0 m high and 3.0 m along the inclined plane. Due to friction the box reaches 4.2 m/s at the bottom of the inclined plane. Beyond the inclined plane lies a spring with 400 N/m constant. It is fixed at its right end. The level ground between the incline and the spring has no friction The box compressed the spring, got pushed back towards the incline by the spring. How far along the inclined plane, from the bottom, will the box temporarily stop on the inclined plane?

A box with 18 kg of mass slides down an inclined plane that is 2.0 m high and 3.0 m along the inclined plane. Due to friction the box reaches 4.2 m/s at the bottom of the inclined plane. Beyond the inclined plane lies a spring with 400 N/m constant. It is fixed at its right end. The level ground between the incline and the spring has no friction The box compressed the spring, got pushed back towards the incline by the spring. How far along the inclined plane, from the bottom, will the box temporarily stop on the inclined plane?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

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. 

Question

**Problem Statement:**

A box with a mass of 18 kg slides down an inclined plane that is 2.0 m high and 3.0 m long along the inclined plane. Due to friction, the box reaches a velocity of 4.2 m/s at the bottom of the inclined plane. Beyond the inclined plane lies a spring with a spring constant of 400 N/m. The spring is fixed at its right end. The level ground between the incline and the spring has no friction.

The box compresses the spring and then gets pushed back towards the incline by the spring. How far along the inclined plane, from the bottom, will the box temporarily stop on the inclined plane?

**Relevant Data:**
1. Mass of the box (m): 18 kg
2. Height of the inclined plane (h): 2.0 m
3. Length of the inclined plane (L): 3.0 m
4. Velocity of the box at the bottom (v): 4.2 m/s
5. Spring constant (k): 400 N/m

**Answer:**

To find the point at which the box will temporarily stop on its way back up the inclined plane, we need to solve for the distance the box travels along the plane after being pushed by the spring. This involves using principles from mechanics such as energy conservation and kinematics, taking into account the forces acting on the box due to gravity, friction, and the restoring force of the spring.

1. Calculate the work done by the spring on the box.
2. Apply energy conservation to relate this work to the gain in potential energy and loss in kinetic energy as the box moves up the incline.

By solving the equations, we can determine the distance along the inclined plane where the box will momentarily come to a stop before potentially sliding back down or remaining at rest depending on the friction. The key equations involve the conservation of mechanical energy, the work-energy theorem, and the understanding of spring dynamics.

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