A 56 g ice cube can slide without friction up a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10.0cm. The spring's constant is 22 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction?|||You can solve this problems by various methods, perhaps work-energy theorem method is the simplest in which path of the object is not important, only initial and final position matters.
After the release of the spring, the potential energy stored in the spring is transferred to the ice cube as kinetic energy.
So, the kinetic energy possessed by the ice cube will be
=(1/2)*(k)*(x^2) where k is spring constant and x is the compression.
=(1/2)*(22)*(0.1^2)
=0.11 J
Now as the cube is rising on the slope, there will come a time when the cube stops and all energy it has will be potential energy by virtue of gravity.
And this potential energy will be equal to its initial kinetic energy.
So, we have
mgh=initial kinetic energy=0.11 J
m=56g=0.056 kg; g=9.8 m/s^2
Solving the equation,
h=0.11/(0.056*9.8)=0.20043 m|||You can solve this problems by various methods, perhaps work-energy theorem method is the simplest in which path of the object is not important, only initial and final position matters.
After the release of the spring, the potential energy stored in the spring is transferred to the ice cube as kinetic energy.
So, the kinetic energy possessed by the ice cube will be
=(1/2)*(k)*(x^2) where k is spring constant and x is the compression.
=(1/2)*(22)*(0.1^2)
=0.11 J
Now as the cube is rising on the slope, there will come a time when the cube stops and all energy it has will be potential energy by virtue of gravity.
And this potential energy will be equal to its initial kinetic energy.
So, we have
mgh=initial kinetic energy=0.11 J
m=56g=0.056 kg; g=9.8 m/s^2
Solving the equation,
h=0.11/(0.056*9.8)=0.20043 m
bst answer
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment