1.20kg mass is dropped from a height of 0.9M unto a spring with force constant K= 120N/m and negligible mass. Find the maximum distance the spring will compressed
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At maximum compression X, the spring's potential energy will equal the gravitational potential energy loss.
M g (H +X) = (k/2)*X^2
Solve for X
To find the maximum distance the spring will compress, we can use the principle of conservation of mechanical energy. The potential energy imparted to the spring is equal to the potential energy initially possessed by the mass.
The potential energy initially possessed by the mass is given by the equation:
PE_initial = m * g * h
where:
m = mass of the object (1.20 kg)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height from which the object is dropped (0.9 m)
Substituting the given values:
PE_initial = 1.20 kg * 9.8 m/s^2 * 0.9 m
PE_initial = 10.584 J (Joules)
Now, the potential energy imparted to the spring when it is compressed is given by:
PE_spring = (1/2) * k * x^2
where:
k = force constant of the spring (120 N/m)
x = maximum distance the spring compresses
Set PE_initial (10.584 J) equal to PE_spring:
10.584 J = (1/2) * 120 N/m * x^2
Solving for x:
x^2 = (2 * 10.584 J) / 120 N/m
x^2 = 0.1764 m^2
Taking the square root of both sides:
x ≈ 0.420 m
Therefore, the maximum distance the spring will compress is approximately 0.420 m.