A block lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 50 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 2.8 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.
What is the blocks position when its kinetic energy is maximum?
Value of Kinetic energy = ?mJ
2007-03-07 05:25:54 · 2 answers · asked by Rogue Bagel 2 in Science & Mathematics ➔ Physics
F=kx, PE=(1/2)kx^2, PE=KE
The block will have the most kinetic energy when the block is at
x=0. At this point all the potential energy in the spring has been converted into kinetic energy because the spring isn't stretched or compressed so no force is being exerted by the spring anymore at this point.
You know that the change in PE plus the change in KE must equal zero by Conservation of Energy Laws and there's no friction.
The equation for energy stored in a spring is (1/2)kx^2. The equation for the expansion of the spring is F=kx so x=F/k or 2.8/50 or 0.056m. So using the other equation the answer for kinetic energy is (1/2)(50)(0.056^2)=0.0784J but your answer is mJ is 78.4mJ.
Your two answers are x=0 and 78.4mJ
2007-03-07 07:07:09 · answer #1 · answered by smartdude474 2 · 0⤊ 0⤋
think of compressing the spring from 0 to a million. At element 0 the rigidity is 0, and at element a million the rigidity is a million. If the rigidity have been a persevering with "a million", then the potential could be Fxd or 1x1 or a million. however the rigidity isn't consistent over that distance. It starts off at 0 and will enhance linearly to a million. So the standard rigidity over that distance is in simple terms a million/2. So the great potential is F(avg) x distance or a million/2. it is the place the a million/2 comes from.
2016-11-23 13:28:43 · answer #2 · answered by guiterrez 4 · 0⤊ 0⤋