Kinetic Energy and Work energy theorem

AI Thread Summary
A toboggan experiences a 47% increase in kinetic energy when a pulling force is applied parallel to the ground. The discussion explores how the kinetic energy change would differ if the pulling force were at a 38° angle above the horizontal. Participants clarify that work is done in both scenarios, resulting in increased kinetic energy, and emphasize the need to express the final velocities in terms of initial velocity, distance, and acceleration. The equations for work done are discussed, with one participant suggesting that the change in work can be calculated using the respective forces and angles. The conversation highlights the importance of understanding the relationship between work and kinetic energy in different force application scenarios.
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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