Answer :
The spring is compressed by 0.122 m before the block comes to rest.
When the block encounters the spring, it will begin to compress the spring due to the force exerted by the spring on the block. This force will cause the block to decelerate and eventually come to a stop when all its kinetic energy has been converted into potential energy stored in the compressed spring.
We can use the conservation of energy principle to determine the compression distance of the spring.
The initial kinetic energy of the block is given by:
[tex]KE_i = (1/2) * m * v^2 = (1/2) * 0.980 kg * (1.32 m/s)^2 = 0.852 J[/tex]
This energy will be converted into potential energy stored in the compressed spring when the block comes to a stop. The potential energy stored in the spring is given by:
[tex]PE_s = (1/2) * k * x^2[/tex]
where k is the force constant of the spring and x is the compression distance of the spring.
Setting the initial kinetic energy equal to the potential energy stored in the spring, we have:
[tex]KE_i = PE_s[/tex]
0.852 J = (1/2) * 245 N/m * x^2
Solving for x, we get:
[tex]x = \sqrt{( (2 * 0.852 J) / (245 N/m) ) } = 0.122 m[/tex]
Therefore, the spring is compressed by 0.122 m before the block comes to rest.
It's important to note that this calculation assumes that the mass of the spring is negligible compared to the mass of the block, and that the spring is ideal with no damping or other losses of energy. In reality, there may be some losses due to friction or other factors, which would result in a smaller compression distance.
For more such questions on spring visit:
https://brainly.com/question/14670501
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