Kinetic Energy and Work energy theorem

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Homework Help Overview

The problem involves a toboggan moving on a horizontal surface with negligible friction, where a pulling force affects its kinetic energy. The original poster seeks to determine the percentage change in kinetic energy when the pulling force is applied at an angle of 38° compared to when it is applied horizontally.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between work done and kinetic energy changes in both scenarios, questioning how to express work when the force is applied at an angle.

Discussion Status

Participants are exploring different expressions for work done in both cases and questioning the adequacy of the given information. Some guidance has been offered regarding assumptions about initial velocity and distance, but no consensus has been reached on how to proceed with the calculations.

Contextual Notes

There is a lack of numerical data provided, leading to uncertainty about calculating the final velocities in both scenarios. Participants are encouraged to assume certain variables for their analyses.

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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