- #1
AmagicalFishy
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Hi, folks. Studying for an upcoming Physics exam. I've got a few example problems which the professor has solved—and am getting a different answer, but I don't know where the fault is in my reasoning. Could someone tell me whether or not my answer is, too, correct?
Find the speed of the block at point C.
Kinetic Energy at Point A = 0
Potential Energy at Point A = Mgh
Kinetic Energy at Point B = [itex]\frac{1}{2}[/itex]MVB2
Potential Energy at Point B = 0
Velocity at Point B = [itex]\sqrt{2gh}[/itex]
Work of Friction = μN(C-B) [μ is the frictional coefficient. N is the normal force. C-B is the distance the block travels on the surface with friction]
I came up with this equation, essentially saying the energy of friction from point B to C taken away from the kinetic energy of the block at point B equals the kinetic energy of the block at point C.
M([itex]\sqrt{2gh}[/itex])2[itex]/[/itex]2 - μN(C-B) = MVc2[itex]/[/itex]2 Which simplifies to:
Mgh - μN(C-B) = MVc2[itex]/[/itex]2
Mgh - μMg(C-B) = MVc2[itex]/[/itex]2
Mg(h-μ)(C-B) = MVc2[itex]/[/itex]2
Solve for Vc to get:
Vc = [itex]\sqrt{2g(h-μ)(C-B)}[/itex]
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The answer my professor gave was: d[itex]\sqrt{k/m}[/itex]
Where d is the point at which the Block compressed the spring by amount d, and k is the spring-coefficient. This answer is much simpler, without a doubt, but I'd like to know if my answer is wrong.
Any help?
Homework Statement
Find the speed of the block at point C.
Homework Equations
Kinetic Energy at Point A = 0
Potential Energy at Point A = Mgh
Kinetic Energy at Point B = [itex]\frac{1}{2}[/itex]MVB2
Potential Energy at Point B = 0
Velocity at Point B = [itex]\sqrt{2gh}[/itex]
Work of Friction = μN(C-B) [μ is the frictional coefficient. N is the normal force. C-B is the distance the block travels on the surface with friction]
The Attempt at a Solution
I came up with this equation, essentially saying the energy of friction from point B to C taken away from the kinetic energy of the block at point B equals the kinetic energy of the block at point C.
M([itex]\sqrt{2gh}[/itex])2[itex]/[/itex]2 - μN(C-B) = MVc2[itex]/[/itex]2 Which simplifies to:
Mgh - μN(C-B) = MVc2[itex]/[/itex]2
Mgh - μMg(C-B) = MVc2[itex]/[/itex]2
Mg(h-μ)(C-B) = MVc2[itex]/[/itex]2
Solve for Vc to get:
Vc = [itex]\sqrt{2g(h-μ)(C-B)}[/itex]
----------
The answer my professor gave was: d[itex]\sqrt{k/m}[/itex]
Where d is the point at which the Block compressed the spring by amount d, and k is the spring-coefficient. This answer is much simpler, without a doubt, but I'd like to know if my answer is wrong.
Any help?
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