|Mar4-12, 03:54 PM||#1|
Work and Changes in Kinetic Energy
1. The problem statement, all variables and given/known data
I. Changes in kinetic energy
A glider moves on a level air track while a hand applies a constant horizontal force. Friction is negligible. The glider starts from rest at point A. The hand continues to push with the same force along the entire length of track.
Describe the motion of the glider. Explain how you determined your answer.
Suppose that the time interval for the glider to move from point A to point B is Δt1. Is the time taken by the glider to mocha from point B to point C greater than, less than, or equal to Δt1? Explain.
Suppose the speed pf the glider increases byΔv1 from point A to point B. Is the increase in speed from point B to point C greater than, less than, or equal to Δv1? Explain how the definition of acceleration can be used to determine your answer.
For one-dimensional motion with constant acceleration, the final speed of an object is related to its displacement Δs by the formulas vf^2 = vi^1 + 2aΔs.
Use this formula to write expressions for the speed of the glider at points B and C in terms of the acceleration of the glider, a, and the distance d.
Use your answer to obtain expressions for the change in speed of the glider between points A and B (ΔVab) and between B and C (ΔVbc).
Is ΔVbc greater than, less than, or equal to ΔVab?
Is your answer consistent with your answer to part A? If not, resolve the inconsistency.
The kinetic energy of an object is defined to be 1/2mv^2.
Use the formula given in part B to derive an equation for the final kinetic energy of an object in terms of the net force on it, the distance Δs that it has moved, and its kinetic energy.
Now express the change in kinetic energy of an object in terms of the net force on the object and the distance Δs that it has moved.
The relationship you have just derived applies to any rigid object moving in one dimension in the direction of a constant net force. State this relationship in words.
2. Relevant equations
PEg = mgH
PEs = 1/2kx^2
|Ff| - |Δx| cos ∅ = Wf - |Δx| |MkN|
3. The attempt at a solution
Increasing velocity from A-D
Less than because force is mass x acceleration so its velocity is increased in this length of track.
Greater than because acceleration is equal to M/S^2
√VE^2 = √2aΔs
Va = √2aΔs
Vb = √0+2ad
Vc = √Vb + 2ad = √2ad + 2ad
Vf^2 = Vi^2 + 2aΔs
2/m aΔs = v^2 - vi^1/ 2 (2/m)
|change in energy, kinetic enegry, speed, work & energy|
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