Kinetic Energy and Work energy theorem

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SUMMARY

The discussion focuses on the application of the work-energy theorem to a toboggan experiencing a change in kinetic energy due to an applied force. When the force is applied parallel to the ground, the kinetic energy increases by 47%. The challenge presented is to determine the percentage change in kinetic energy when the force is applied at an angle of 38° above the horizontal. Key equations referenced include the work equation \( W = Fd \cos \theta \) and the kinetic energy equation \( KE = \frac{1}{2}mv^2 \).

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jasonbans
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Homework Statement


A toboggan is initially moving at a constant velocity along a snowy horizontal
surface where friction is negligible. When a pulling force is applied parallel to
the ground over a certain distance, the kinetic energy increases by 47%. By what
percentage would the kinetic energy have changed if the pulling force had been
at an angle of 38° above the horizontal?


Homework Equations



w= Fd cos theta
w= 1/2mv^2

The Attempt at a Solution


so Ek = 1.47 Ek intial Ek = kinetic energy
 
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37%?
 
yeah can you explain what you did?
 
First find an expression for the change in KE when the force is applied along the horizontal.
 
you mean a expression for both cases?
we know that in the first case the force and displacement act in the same direction meaning no work is done.
and the second case i don't get, how can you get W = (Fcos38)d where F and d is not even given
 
jasonbans said:
we know that in the first case the force and displacement act in the same direction meaning no work is done.
and the second case i don't get, how can you get W = (Fcos38)d where F and d is not even given

Work is done in BOTH cases. This work results in an increase in KE.
No more numerical data is necessary. Just assume initial velocity is u in both cases and final is V for the first case and v for the second (symbols chosen so that big V is for the greater value and small v for the smaller value - I hate to call them v1 and v2!) since the final will be different for the two cases.
 
how am i suppose to find the velocity if they give so little information?
 
Assume that applied force is F in first case and Fcos38 in the second case.
Initial velocity is u in both.
Distance traveled is assumed to be constant e.g. x in both.
 
What is V in terms of u, x and acceleration?
 
  • #10
grzz said:
What is V in terms of u, x and acceleration?

i did this instead , this should make sense right? cause the change in W = the final kinetic energy - the intial kinetic energy, and W = Fd in first case and second is W=fdcos38
 

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  • #11
I think it is OK.
 

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