### Problem 2A spring (k = 75.0 N/m) has a natural length of 1.00 m. The spring is compressed to a length of 0.500 m and a block of mass 1.50 kg is attached at its free end on a frictionless slope which makes an angle of 41.0° with respect to the horizontal as shown in the figure below. The spring is then released from rest. **How far up the slope from the starting position will the block move up the incline before coming to rest?**#### Diagram ExplanationIn the provided diagram, a spring is compressed and attached to a block on a frictionless incline. The incline makes an angle of 41° with the horizontal.- The compressed length of the spring is labeled as 0.50 m.- The incline is clearly marked with the angle θ = 41°.- A horizontal line at the bottom of the diagram indicates a length of 2m.Below the diagram, an additional question is presented:#### Additional QuestionThe incline is cut so that the length of the incline is now 1 m. **How much should the spring be compressed if the block needs to hit an object 2 m away?**---Graphical or computational steps would typically follow to solve for the potential energy stored in the compressed spring, equate it to the gravitational potential energy when the block moves up the slope, and solve for the distance traveled or the new compression length of the spring.

Answered: Image /qna-images/answer/6e4436b4-8e21-4753-87c7-b20d8ed061a9.jpg

A spring (k = 75.0 N/m) has a natural length of 1.00 m. The spring is compressed to a length of 0.500 m and a block of mass 1.50 kg is attached at its free end on a frictionless slope which makes an angle of 41.0° with respect to the horizontal as shown in the figure below. The spring is then released from rest. How far up the slope from the starting position will the block move up the incline before coming to rest? 0,50 m e =41° 2m The incline is cut so that the length of the incline is now 1 m. How much should the spring be compressed if the block needs to hit an object 2 m away?

A spring (k = 75.0 N/m) has a natural length of 1.00 m. The spring is compressed to a length of 0.500 m and a block of mass 1.50 kg is attached at its free end on a frictionless slope which makes an angle of 41.0° with respect to the horizontal as shown in the figure below. The spring is then released from rest. How far up the slope from the starting position will the block move up the incline before coming to rest? 0,50 m e =41° 2m The incline is cut so that the length of the incline is now 1 m. How much should the spring be compressed if the block needs to hit an object 2 m away?

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 2

A spring (k = 75.0 N/m) has a natural length of 1.00 m. The spring is compressed to a length of 0.500 m and a block of mass 1.50 kg is attached at its free end on a frictionless slope which makes an angle of 41.0° with respect to the horizontal as shown in the figure below. The spring is then released from rest. **How far up the slope from the starting position will the block move up the incline before coming to rest?**

#### Diagram Explanation

In the provided diagram, a spring is compressed and attached to a block on a frictionless incline. The incline makes an angle of 41° with the horizontal.

- The compressed length of the spring is labeled as 0.50 m.
- The incline is clearly marked with the angle θ = 41°.
- A horizontal line at the bottom of the diagram indicates a length of 2m.

Below the diagram, an additional question is presented:

#### Additional Question

The incline is cut so that the length of the incline is now 1 m. **How much should the spring be compressed if the block needs to hit an object 2 m away?**

---

Graphical or computational steps would typically follow to solve for the potential energy stored in the compressed spring, equate it to the gravitational potential energy when the block moves up the slope, and solve for the distance traveled or the new compression length of the spring.

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