Springy question....

[xcutexboax]

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Hi Guys... came across this qns...may seem quite simple but it is actually quite tricky.

A Block of mass m=1.00kg situated on an incline at an angle of 43.0 is connected to a spring of negligible mass having a spring constant of 100N/m. the pulley and the incline are frictionless. The block is released from rest wif the spring initially unstretched.

How far does it move down the incline before coming to rest???
What is its acceleration at its lowest point??


can i jus equate the work done in extension of spring: kx^2/2 equals to (mgsin43.0) x (x)??? Pls enlighten me...

and for acceleration... at its lowest point... do i make use of 1/2mv^2?? from the work done? then using v^2=U^2 +2as to determine acceleration?

I personally feel acceleration shd be constant... cos of negligible friction..Don't really know if i am correct... Pls enlighten me.. :confused: Thanks :smile:

[xcutexboax]

Woooh... thanks i finally solve it....

an additional part of the question:: Consider that when the block comes to rest, the change in gravitational potential energy is converted to the elastic potential energy of the spring. However, there is still a net force exerted on the block at that moment.

Simply jus equate the work done in extension to mgsin30 x extension.

and then use normal resolving of forces = ma since there is a net force... thanks guys. :cool:

[Tom McCurdy]

hahaha springs 1/2kx^2 = mgy those were really fun problems to do in physics one