Rate of spring compession from block on incline

In summary: Block 1 has a mass of 10 kg and Block 2 has a mass of 1 kg. The spring constant is 500 N/kg. The angle is 35 degrees. The acceleration due to gravity is 9.8 m/s2. The ukf is 0.1. In summary, the block 1 slides down an slope to hit block 2, which compresses the spring. The spring constant is 500 N/kg, the angle is 35 degrees, and the acceleration due to gravity is 9.8 m/s2. The block 1 has a mass of 10 kg and the block 2 has a mass of 1 kg. The collision between the blocks is elastic. The velo of block 1 just before hitting block 2
  • #1
december7
3
0

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)
 
Physics news on Phys.org
  • #2
Not enough information. What are the initial conditions for Block 1? How long before it hits Block 2?
 
  • #3
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.
 
  • #4
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 ...
 
  • #5
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.
 
  • #6
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.
 

FAQ: Rate of spring compession from block on incline

What is the formula for calculating the rate of spring compression from a block on an incline?

The formula for calculating the rate of spring compression from a block on an incline is: k = (mgsinθ - μmgcosθ) / x, where k is the spring constant, m is the mass of the block, g is the acceleration due to gravity, θ is the angle of the incline, μ is the coefficient of friction, and x is the distance the spring is compressed.

How does the angle of the incline affect the rate of spring compression?

The angle of the incline affects the rate of spring compression by changing the component of the force of gravity acting on the block and the frictional force. As the incline becomes steeper, the force of gravity acting on the block increases, resulting in a higher rate of spring compression. Similarly, an increase in the coefficient of friction will also lead to a higher rate of spring compression.

What is the relationship between mass and the rate of spring compression?

The relationship between mass and the rate of spring compression is inverse. As the mass of the block increases, the rate of spring compression decreases. This is because a heavier block will require more force to move up the incline, resulting in a smaller distance the spring is compressed.

How does the spring constant affect the rate of spring compression?

The spring constant directly affects the rate of spring compression. A higher spring constant means that the spring is stiffer and will require more force to compress it a certain distance. As a result, a higher spring constant will lead to a lower rate of spring compression.

Does the rate of spring compression change as the spring is compressed further?

Yes, the rate of spring compression changes as the spring is compressed further. This is due to the fact that the spring constant is not constant and varies with the displacement of the spring. As the spring is compressed further, the spring constant increases, resulting in a lower rate of spring compression.

Back
Top