The speed of the block when it is 10 cm from the equilibrium position is 3.87 m/s.
We need to apply the concept of conservation of energy. The conservation of the initial kinetic energy, the final kinetic energy, and the elastic potential energy must be applied in this situation. Mathematically this can be expressed as,
KE i = KE f + PE f
1/2* m* vi² = 1/2* m* vf² + 1/2* k * x²
where,
m is mass
vi is initial velocity
vf is final velocity
k is spring constant
x is displacement
Given that, m = 2 kg
k = 200 N/m
x = 10 cm = 0.1 m
vi = 4 m/s
Placing them in the above equation, we have
1/2* 2* 4² = 1/2* 2* vf² + 1/2* 200* 0.1²
16 - 1 = vf²
vf² = 15
vf = 3.87 m/s
Thus, the speed of the block when it is 10 cm from the equilibrium position is 3.87 m/s.
The question is incomplete. the complete question is 'what is its speed when it is a distance of 10 cm from the equilibrium position?'
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