Energy Problem involving Friction and Spring

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Homework Help Overview

The problem involves a wooden block projected up an incline by a compressed spring, with considerations of kinetic friction and gravitational potential energy. The scenario is set on a slope inclined at 35 degrees, and the participants are tasked with calculating the potential energy stored in the spring.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss energy conservation principles, including kinetic energy, potential energy, and work done against friction. There is an attempt to set up an energy equation, but some participants express uncertainty about the inclusion of potential energy and the correct terms to use.

Discussion Status

The discussion is ongoing, with participants questioning the completeness of the energy equation and the inclusion of potential energy. Some guidance has been offered regarding the need for additional terms in the energy equation, indicating a productive exploration of the problem.

Contextual Notes

Participants are navigating the complexities of energy transformations and the effects of friction, with some uncertainty about the definitions and terms involved in the energy equation.

Chandasouk
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Homework Statement



A wooden block with mass 1.30 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 35.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 4.15 m up the incline from A, the block is moving up the incline at a speed of 5.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is [tex]\mu[/tex]k = 0.450. The mass of the spring is negligible.Calculate the amount of potential energy that was initially stored in the spring.
Take free fall acceleration to be g = 9.80 m/s^2.

I don't really know what to do.

I start by doing some energy graphs
(Initial) Us = KE + [tex]\Delta[/tex]Uint (Final State)

1/2K[tex]\Delta[/tex]x2 = 1/2mv2+[tex]\mu[/tex]mgcos[tex]\theta[/tex]*L

L being 4.15m
 
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In the energy equation you have not included the potential energy.
 
PE is included because the block is at a location higher than it's initial position right?
 
What is the final position of the block from the horizontal surface?
ΔUint includes only the work done against the frictional force.
So one more term must be there in the energy equation.
 

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