How to Calculate the Compression Distance of a Spring on an Inclined Plane?

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Homework Help Overview

The problem involves a block sliding down an inclined plane towards a spring, with the goal of determining the compression distance of the spring when the block comes to rest. The scenario includes parameters such as the angle of the incline, the mass of the block, the spring constant, and the initial speed of the block.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning the correct setup of the energy equations. There is an exploration of the relationship between the distance traveled by the block and the compression of the spring.

Discussion Status

Some participants have provided guidance on using energy equations, while others are exploring the implications of the block's motion and its acceleration. There is an ongoing examination of the assumptions made regarding the distances involved and the forces acting on the block.

Contextual Notes

Participants note that the block travels a distance greater than the initial distance from the spring due to the compression, raising questions about how to accurately account for this in their calculations. There is also a mention of potential misunderstandings in the explanations provided.

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Homework Statement



an inclined plane of angle 20 degree has a spring of force constant k= 500N/m fastened securely at the bottom so that the spring is parallel to the surface. a block of mass m=2.5kg is placed on the plane at a distance d=0.300m from the spring. From this position, the block is projected downward towards the spring with speed, v=0.750m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Homework Equations



work done of spring=1/2kx2 and kinetic energy=1/2mv2
and potential energy=mglsin20?

The Attempt at a Solution


i try to solve the ques by using work done of spring= potential energy+kinetic energy
bt can't find the correct ans =S
which part i did wrong?
 
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Show your work in detail.

ehild
 
ehild said:
Show your work in detail.

ehild

1/2mv2+mglsin20=1/2kx2
1/2(2.5)(0.75)^2+2.5(9.8)(0.3sin20)=1/2(500)(x)^2
x=0.113m
bt the real ans is o.131m
i wonder which part i did wrong
 
The block travels more than 0.3 m as its compresses the spring by x.

ehild
 
ehild said:
The block travels more than 0.3 m as its compresses the spring by x.

ehild

ic...so how do i find the distance traveled by the block?
i got to find the acceleration 1st rite?
thru mgsin20=ma?
 
Use the energy equation as before but with the new length the block travels till coming to rest.

ehild
 
ehild said:
Use the energy equation as before but with the new length the block travels till coming to rest.

ehild

i mean how do find the new length the block travelled?
is the block's acceleration-----mgsin20=ma?
 
Before touching the spring, it is. After touching the spring, it is different. But you do not need the acceleration. Think: what is x? Make a drawing. ehild
 
ehild said:
Before touching the spring, it is. After touching the spring, it is different. But you do not need the acceleration. Think: what is x? Make a drawing.


ehild

lolx..silly me XDD
i missunderstand ur explanation jus now
i figured it out..
thx for ur help ^^
 

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