Rate of spring compession from block on incline

AI Thread Summary
The discussion centers on calculating the compression of a spring when block 1, weighing 10 kg, slides down a 35-degree incline and collides with block 2, which weighs 1 kg. Key equations involve the work done on the spring and the forces acting on block 1, including gravitational force and friction. Participants suggest using a free body diagram to clarify the forces and determine the velocity of block 1 before impact. The effect of block 2 on the spring compression is debated, particularly whether the collision is elastic or inelastic. Understanding these dynamics is essential for accurately calculating the spring's compression.
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Homework Statement



Block 1 slides down an slope to hit block 2, which compresses a spring. How much compression does the spring experience? Mass - block 1=10 kg, block 2 = 1 kg, spring constant = 500 N, angle = 35, acceleration due to gravity=9.8
m/s, ukf = 0.1

Homework Equations



ws = (1/2)*kx^2 = w(net force) d cos 0
ws= (1/2)*500*x^2 = (9.8*10) 3 (cos 35)


The Attempt at a Solution



without the coefficient of static, and the mass of block 2
ws= (1/2)*500*.9815^2 = (9.8*10)3(cos35)
1=240.83
not sure how to work in the coefficient of static on block 1, and uncertain how to account for the effect of block 2 (1 kg)
 
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Not enough information. What are the initial conditions for Block 1? How long before it hits Block 2?
 
Block 1 is at rest, distance is 3M to Block 2.
I am unsure if the the '(9.8*10) 3 (cos 35)'
can be used, as well as how to integrate the
coefficient of 0.1 for ufk. Finally, how to deal
with this force when it meets block 2 (1 kg)
which separates Block 1 from the spring.
Thanks.
 
Best way to get the Block 1 situation clear is to draw a free body diagram. That will give you the force accelerating it (the component of its weight acting down the slope less frictional force). That will lead to knowing how fast it is traveling when it hits Block 2 ...
 
I have calculated 10*9.8*sin35 = 56.21 - 10*9.8*cos35*0.1 = 48.18 * 3M
= 144.54N. This is the only way I have been able to find to do this. I'm not sure how to deal with the 1 kg block which sits at the end of the spring. The calculated mass will be exerted against the 1 kg block, which in turn will compress the spring X amount. It is this amount of compression that is sought. Based on PES = (1/2)KX^2. (with spring K=500) Forgetting about the 1 kg block for a moment, I thought I could possibly take 144.50/250 =0.578 = square root of 0.578 = 0.76M spring compression. Even if this is close to correct, it disregards the 1 kg block. Any input appreciated. Thanks.
 
You also have to know if the collision between the blocks is elastic or not. That way, you’ll know how much energy is transferred to block 2.

The velo of block 1 just before hitting block 2 should be clear enough, since all the data are given.
 
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