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Old Dec 3rd 2009, 05:42 PM   #1
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Find speed of block on a ramp/pulley

A 20kg block (m1) slides on a 40 degree frictionless ramp. This block is connected to a 30kg block (m2) by a string that passes over a frictionless pulley. The 30kg block is 20cm above the floor and connected to a spring of negligible mass with spring constant 250N/m. The spring is initially unstretched and connected to the floor. The 20kg block is pulled a distance of 20cm down the ramp (so that the 30kg block is 40cm above the floor) and is released from rest. Calculate the speed of each block when the 30kg block is again 20cm above the floor (spring unstretched).

My attempt:
I assumed that you had to use conservation of energy to find v:
Usi + Ugi +KEi = Usf + Ugf + KEf,
where Us is Spring potential, Ug is gravitational potential and KE is kinetic energy.

KE is initially zero so we have
0.5k(xi)^2 + mghi = 0.5k(xf)^2 + mghf + 0.5mv^2

xi is 0.2m, xf is 0.4m for block 2.
I worked everything out and got 10v^2 = -64...which I'm pretty sure is incorrect. Can anyone guide me to the right direction??

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Old Dec 4th 2009, 01:18 AM   #2
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Let us first look at the problem in the absence of the spring.

Let a be the accln of the system and T be the tension in the string

We have for the 20 kg block

T - 20 g sin 40 = 20 a and

30 g - T = 30 a for the 30 kg block.

Eliminating T we find that the accln a = 3.36 m/sec.

Without the spring, the motion of the 30 kg block would be identical to that on a planet where g = 3.36 m/s if it were falling freely.

Let the potential be zero at 20 cm above the floor.

When the block is pulled up by 20 cm its effective increase in grav P.E. on that planet will be 30 x 3.36 x 0.02.

But in fact it is connected to a spring, and the extension results in an additional P.E. of 0.5 x 250 x (0.02)^2.

Thus the total P.E. is

30 x 3.36 x 0.02 + 0.5 x 250 x (0.02)^2 = 2.066.

from conservation of energy, this must be equal to the K.E. when it reaches back to 20 cm above the floor. Hence

0.5 x 30 x v^2 = 2.066 giving v = 0.371 m/s or 37.1 cm/sec.

Is the answer OK ?
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