Kinetic Energy and Work energy theorem

In summary: However, it can be simplified. Since the distance, initial velocity, and mass are all the same, the only difference between the two cases is the applied force. So we can just say that the kinetic energy increases by a factor of 1.47 in the first case and a factor of something else in the second case.In summary, the conversation discusses the change in kinetic energy when a pulling force is applied to a toboggan initially moving at a constant velocity on a frictionless surface. It is determined that the kinetic energy will increase by 47% when the force is applied parallel to the ground. The conversation then explores what the change in kinetic energy would be if the force was applied at an angle of 38° above the
  • #1
jasonbans
45
0

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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  • #2
37%?
 
  • #3
yeah can you explain what you did?
 
  • #4
First find an expression for the change in KE when the force is applied along the horizontal.
 
  • #5
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
 
  • #6
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.
 
  • #7
how am i suppose to find the velocity if they give so little information?
 
  • #8
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.
 
  • #9
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.
 

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is dependent on the mass and velocity of the object.

How is kinetic energy calculated?

Kinetic energy is calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

What is the work energy theorem?

The work energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In other words, the work done on an object will result in a change in its speed or direction of motion.

What is the relationship between kinetic energy and work?

Kinetic energy and work are related because work is the transfer of energy from one object to another. When work is done on an object, its kinetic energy changes.

Can kinetic energy be converted into other forms of energy?

Yes, kinetic energy can be converted into other forms of energy such as potential energy, thermal energy, and sound energy. This is known as the principle of conservation of energy.

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