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

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To calculate the compression distance of a spring on an inclined plane, the problem involves a block of mass 2.5 kg sliding down a 20-degree incline toward a spring with a force constant of 500 N/m. The block is projected with an initial speed of 0.750 m/s from a distance of 0.300 m from the spring. The energy conservation equation combines kinetic energy, potential energy, and the work done by the spring, leading to the calculation of compression distance. The initial attempt yielded an incorrect compression distance, prompting a reevaluation of the distance traveled by the block before it comes to rest. Ultimately, the correct compression distance is determined to be 0.131 m after clarifying the energy considerations involved.
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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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