anonymous
  • anonymous
A 13kg box slides 4.0m down the frictionless ramp shown in the figure , then collides with a spring whose spring constant is 170 N/m. Figure : http://i.imgur.com/0M7us.png What is the maximum compression of the spring? At what compression of the spring does the box have its maximum speed? I'm stuck on the first question. I thought I use mgh = (1/2)kx^2? so, (13kg)(9.8m/s^2)(4sin(30)) = (1/2)(170N/m)x^2. x = √(2mgh / k) = 1.7m? But it's telling me this is wrong Any help please?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
We need to be careful in our expression for potential energy of the box at the top of the ramp. Since gravity is vectored along the ramp, let's use the Work-Energy theorem. \[W = F \cdot d = \Delta KE \rightarrow mg \sin(\theta) d = {1 \over 2} mv^2\] We can then use the following expression\[mg \sin(\theta)d = {1 \over 2} kx^2\]
anonymous
  • anonymous
I don't understand, I have to find v first?
anonymous
  • anonymous
Nah. Just realize that the work done by the gravitational force equals a change in kinetic energy. Then all of this energy is stored in the spring.

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anonymous
  • anonymous
always remwmber that energy gets transferred from one source to the next(if no external forces act) so the question is how in this case from kinetic energy to p.e of sping use work-energy theorem to get chnge in k.e=net work done on the body here only 1 force(gravity) acts on the body and wat is that work done =? so this k.e gets converted to p.e of spring=1/2kx^2=work done by restoring force -kx remeber the first point always
mos1635
  • mos1635
i think i see your mistake :) |dw:1333038128640:dw| note!!!! total slide=4m + x !!! there fore h=(4+x) sin30..............
mos1635
  • mos1635
|dw:1333038381149:dw|

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