A massless spring with spring constant 23 N/m hangs vertically. A body of mass 0.85 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find how far below the initial position the body descends.
I got .362 m, but thats not right and i'm stuck on how to solve it, thanks for the help. (:
I got .362 m, but thats not right and i'm stuck on how to solve it, thanks for the help. (:
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The value ".362 meters" is the distance at which the net force is zero, (i.e. where the upward pull of the spring matches the weight of the body); also called the "equilibrium" position. But that is NOT the lowest point that the body reaches.
This is because you are releasing the body from rest and allowing it to fall. As it gains momentum, it falls PAST the equilbrium position and continues to descend, reaching a point that is LOWER than the equilibrium position. It ends up oscillating around the equilibrium position.
One way to analyze this is in terms of conservation of energy (assuming no losses due to friction). In this case, we have:
Total energy at top = total energy at bottom
The total energy at any given point consists of the sum of KE, gravitaional PE, and elastic PE at at that point. If we say that the body falls a distance "h" from top to bottom, and we say that its GPE is zero at the bottom, then we have:
Energy at top:
KE + GPE + EPE
0 + mgh + 0
Energy at bottom:
KE + GPE + EPE
0 + 0 + ½kh²
Set these two equal:
0 + mgh + 0 = 0 + 0 + ½kh²
Solve for "h":
h = 2mg/k
This is TWICE the value you had. (This makes sense, because as it oscillates, it falls as far below the equilibrium position as it originally was above it).
This is because you are releasing the body from rest and allowing it to fall. As it gains momentum, it falls PAST the equilbrium position and continues to descend, reaching a point that is LOWER than the equilibrium position. It ends up oscillating around the equilibrium position.
One way to analyze this is in terms of conservation of energy (assuming no losses due to friction). In this case, we have:
Total energy at top = total energy at bottom
The total energy at any given point consists of the sum of KE, gravitaional PE, and elastic PE at at that point. If we say that the body falls a distance "h" from top to bottom, and we say that its GPE is zero at the bottom, then we have:
Energy at top:
KE + GPE + EPE
0 + mgh + 0
Energy at bottom:
KE + GPE + EPE
0 + 0 + ½kh²
Set these two equal:
0 + mgh + 0 = 0 + 0 + ½kh²
Solve for "h":
h = 2mg/k
This is TWICE the value you had. (This makes sense, because as it oscillates, it falls as far below the equilibrium position as it originally was above it).
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F = -KX
X = -F / K
X = -(0.85)(-9.8) / (23)
X = 0.36 m <----------Answer
X = -F / K
X = -(0.85)(-9.8) / (23)
X = 0.36 m <----------Answer