Spring Potential energy, should be easy?

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Homework Help Overview

The problem involves a spring with a spring constant of 500 N/m used to launch a 0.42-kg mass up an inclined plane. The mass is propelled after the spring is compressed by 30 cm, and the scenario includes considerations of kinetic friction on both the horizontal surface and the inclined plane.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between elastic potential energy and kinetic energy, questioning the impact of friction on the calculations. There are attempts to clarify the correct compression distance of the spring and how to incorporate friction into the energy balance.

Discussion Status

Some participants have provided guidance on calculating the work done by friction and its effect on the overall energy balance. There is ongoing exploration of the correct approach to solving for the mass's speed at different points in its motion, with some participants expressing uncertainty about the initial assumptions made in their calculations.

Contextual Notes

Participants note the importance of accurately determining the distance over which friction acts and the need to consider the negative work done by friction in the energy equations. There is a recognition of the initial misunderstanding regarding the spring compression distance.

Patdon10
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Homework Statement



A spring with a spring constant of 500 N/m is used to propel a 0.42-kg mass up an inclined plane. The spring is compressed 30 cm from its equilibrium position and launches the mass from rest across a horizontal surface and onto the plane. The plane has a length l = 3 m and is inclined at 32°. Both the plane and the horizontal surface have a coefficient of kinetic friction with the mass of 0.35. When the spring is compressed, the mass is at distance d = 1.6 m from the bottom of the plane.

6-p-057-alt.gif



(a) What is the speed of the mass as it reaches the bottom of the plane?
1Your answer is incorrect.

(b) What is the speed of the mass as it reaches the top of the plane?
2

(c) What is the total work done by friction from the beginning to the end of the mass's motion?


Homework Equations



E = K + PE
Ki + PEi = Kf PEf
Elastic Potential Energy = (1/2)kx2

The Attempt at a Solution



I'm pretty sure I can solve this fairly easily after I figure out part A. However, Part A is giving me a lot of trouble.

K_i + Elastic PE_i = K_f + Elastic PE_f
0 + (1/2)(500)(1.62) = (1/2)(0.42)(v2) + 0
v = 55.205 m/s

That's wrong, any advice?

Thanks in advance.
 
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Patdon10 said:
K_i + Elastic PE_i = K_f + Elastic PE_f
0 + (1/2)(500)(1.62) = (1/2)(0.42)(v2) + 0
v = 55.205 m/s
(1) You forgot about friction! (What's the work done by friction?)
(2) The compression of the spring is not 1.6 m.
 
Doc Al said:
(1) You forgot about friction! (What's the work done by friction?)
(2) The compression of the spring is not 1.6 m.

The work done by friction is the Frictional force multiplied by the distance. Would calculating that work value help me find the velocity. I remembered I had a bigger problem trying to figure out this problem that I forgot to include in the 'attempt at a solution'.

This is what I've got:
I have an initial elastic potential energy, and a final kinetic energy. With the givens, I can also calculate the force of friction on the block, but how do I incorporate that into the equation?
Even if I can find a force from the spring, I could use Newton's 2nd law to find the final speed.

Any insight?
 
Patdon10 said:
I have an initial elastic potential energy, and a final kinetic energy. With the givens, I can also calculate the force of friction on the block, but how do I incorporate that into the equation?
Initial mechanical energy - work done by friction = final mechanical energy

The work done by friction is negative work: it decreases the mechanical energy.

What's the work done by friction? Incorporate that into your energy balance as I indicate above.
 
Ok, so let's try this again:
Work done by friciton is:
F_fr = mgμ = (0.42)(9.81)(0.35) = 1.44 N
W_fr = (1.44N)(1.6m) = 2.307 Nm

0 + (1/2)(500N/m)(0.3m^2) - 2.307 Nm = (1/2)(0.42)(v^2) + 0
22.5 - 2.307 = 0.21v^2
v = 9.806 m/s

Which is exactly right! Thank you so much. I wish I would have thought to put it into a work equation. I need to pay more attention to the units to see that I can do things like that. From here on out, I should be able to just use kinematics
 

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