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Kiwigami
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New user has been reminded to post schoolwork in the Homework Help forums and fill out the HH Templatge
Here is a made up function of force changing with respect to displacement: F(x) = 2000 - 100x
The force is applied onto an object, pushing it horizontally. Ignore friction and air resistance. The mass is constant. I'm trying to find Velocity with respect to displacement.
Attempt 1:
Knowing F = m*a:
2000 - 100x = m*a
a(x) = (2000-100x)/m
To find velocity, I take the integral of a(x) to get v(x) = [-100(0.5x2 - 20x)]/m
Attempt 2:
Now, I believe that Work = Force * Displacement. W = ∫ from 0 to b of (2000 - 100x)dx = [-50(b-40)b]
Since the object is only sliding horizontally, there's no potential energy, but there is kinetic energy. I set [-100(0.5x2 - 20x)] = 0.5mv^2 and solve for v:
v(x) = sqrt[(2/m) * [-50(x-40)x]]
In the end, I get two different answers. What I would like to know is what am I doing wrong? Did I use illegal math? Did I get the formulas wrong? Did I use the forumlas wrong?
The force is applied onto an object, pushing it horizontally. Ignore friction and air resistance. The mass is constant. I'm trying to find Velocity with respect to displacement.
Attempt 1:
Knowing F = m*a:
2000 - 100x = m*a
a(x) = (2000-100x)/m
To find velocity, I take the integral of a(x) to get v(x) = [-100(0.5x2 - 20x)]/m
Attempt 2:
Now, I believe that Work = Force * Displacement. W = ∫ from 0 to b of (2000 - 100x)dx = [-50(b-40)b]
Since the object is only sliding horizontally, there's no potential energy, but there is kinetic energy. I set [-100(0.5x2 - 20x)] = 0.5mv^2 and solve for v:
v(x) = sqrt[(2/m) * [-50(x-40)x]]
In the end, I get two different answers. What I would like to know is what am I doing wrong? Did I use illegal math? Did I get the formulas wrong? Did I use the forumlas wrong?
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