Solve Inclined Plane System with Spring & Mass

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An inclined plane with a 20-degree angle has a spring with a force constant of 500 N/m at its base, and a 2.50 kg block is projected downwards towards the spring at 0.750 m/s from a distance of 0.300 m. The problem requires determining the distance the spring compresses when the block momentarily comes to rest. The initial approach using force equations is critiqued, emphasizing that the system is not in static equilibrium due to the block's acceleration. A recommendation is made to utilize conservation of energy for a more straightforward solution to find the spring's compression distance. The book states the answer is 0.131 m, but discrepancies in calculations are noted.
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Homework Statement



an inclined plane of angle 20 degrees has aspring of force constant k=500 N/m fastened securely at the bottomso that the spring is parallel to the surface. a block of massm=2.50kg is placed on the plane at a distance d=0.300m from thespring. from this position, the block is projected downward towardthe spring with a speed v=0.750 m/s. by what distance is the springcompressed when the block momentarily comes to rest?

Homework Equations



F = kx and F = ma

The Attempt at a Solution



The book says the answer is 0.131 m. But I get different answer. Here's what I'm doing.
a = g * sinθ

m * ax = k * x

Can someone please help me??
 
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Pyae said:
The book says the answer is 0.131 m. But I get different answer. Here's what I'm doing.
a = g * sinθ

m * ax = k * x
Your attempted approach might work if the system was in a state of static equilibrium, but it's not. Even though the instantaneous velocity of the block is zero, it is accelerating and that makes a big difference.

I suggest using conservation of energy to solve this problem (you could use kinematics to solve it, but I think conservation of energy will be a little easier).
 
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