- #1
jace1313
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so here is the problem
The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 40 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.7 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.a) What is the position of the block?
So my original thinking was simplistic, we have two horizontal forces and the block will stop when equilibrium is achieved (when the force applied equals force of the spring):
2.7 = kx
2.7 = 40x
x = 0.0675
Solution teacher gave:
Work done by force = final potential energy of spring
F*x = 1/2kx^2
2.7*x = 1/2(40)x^2
x = 0.135
So I understand this thinking as well, the work done by the force puts energy into the spring (potential energy). But if this method works, shouldn't the force of the spring and the force applied equal each other in the end?
If we calculate the force of the spring (40*0.135) it is 5.4 N and we are given the force applied as 2.7 N. The net force on the block would be 5.4 - 2.7, which is 2.7 N. And if there is a net force on the block, the block is accelerating and not in a final resting position.
Is it just that the block stops temporarily in that position, but will eventually rest in the equilibrium I solved for?
The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 40 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.7 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.a) What is the position of the block?
So my original thinking was simplistic, we have two horizontal forces and the block will stop when equilibrium is achieved (when the force applied equals force of the spring):
2.7 = kx
2.7 = 40x
x = 0.0675
Solution teacher gave:
Work done by force = final potential energy of spring
F*x = 1/2kx^2
2.7*x = 1/2(40)x^2
x = 0.135
So I understand this thinking as well, the work done by the force puts energy into the spring (potential energy). But if this method works, shouldn't the force of the spring and the force applied equal each other in the end?
If we calculate the force of the spring (40*0.135) it is 5.4 N and we are given the force applied as 2.7 N. The net force on the block would be 5.4 - 2.7, which is 2.7 N. And if there is a net force on the block, the block is accelerating and not in a final resting position.
Is it just that the block stops temporarily in that position, but will eventually rest in the equilibrium I solved for?