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Change in Kinetic Energy/ Work

  • Thread starter Jrlinton
  • Start date
  • #1
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Homework Statement


000000000.PNG


Homework Equations


ΔKE=W+FcosΘd
KE=.5mv^2
Final Vel^2=initial Velocity^2+2ax

The Attempt at a Solution


So calculate the force at x=2.6 to -5.356N, set 2.6(-5.356)=1.3Vf^2-1.3(0^2) which seemed to prove a problem as the square of the final velocity was negative, but just used that negative square in the difference of kinetic energy equation and got -5.356 J- 0= -5.356 as the change in Kinetic energy or work. I am skeptical of my methods and also a little clueless on the second part of the problem.

Many thanks
 

Answers and Replies

  • #2
SammyS
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Homework Statement


View attachment 107157

Homework Equations


ΔKE=W+FcosΘd
KE=.5mv^2
Final Vel^2=initial Velocity^2+2ax

The Attempt at a Solution


So calculate the force at x=2.6 to -5.356N, set 2.6(-5.356)=1.3Vf^2-1.3(0^2) which seemed to prove a problem as the square of the final velocity was negative, but just used that negative square in the difference of kinetic energy equation and got -5.356 J- 0= -5.356 as the change in Kinetic energy or work. I am skeptical of my methods and also a little clueless on the second part of the problem.

Many thanks
Please display the problem statement in a form that can be read with reasonable effort.

Perhaps, type it out.
 
  • #3
134
1
A 2.6 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529127entrance1_N10035.mml?size=14&ver=1475946606896&algorithm=1&rnd=1475946607102 [Broken], where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.6 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.6 m?
 
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  • #4
SammyS
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A 2.6 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2529127entrance1_N10035.mml?size=14&ver=1475946606896&algorithm=1&rnd=1475946607102 [Broken], where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.6 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.6 m?
That's better. We also have:
F(x) = (4.7 - x2)i newtons.
or
##\ \vec F (x)=(4.7-x^2)\hat i \ ##​

What is F at x = 0? What is F at other locations?
 
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  • #5
Ray Vickson
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Homework Statement


View attachment 107157

Homework Equations


ΔKE=W+FcosΘd
KE=.5mv^2
Final Vel^2=initial Velocity^2+2ax

The Attempt at a Solution


So calculate the force at x=2.6 to -5.356N, set 2.6(-5.356)=1.3Vf^2-1.3(0^2) which seemed to prove a problem as the square of the final velocity was negative, but just used that negative square in the difference of kinetic energy equation and got -5.356 J- 0= -5.356 as the change in Kinetic energy or work. I am skeptical of my methods and also a little clueless on the second part of the problem.

Many thanks
Your formula ##V_f^2 = V_i^2 + 2ax## does not work; it assumes that ##a## is a constant, which is not true in this problem.
 
  • #6
134
1
Okay so I integrated the force to 4.7x-x^3/3. I am unsure of what to do next
 
  • #7
134
1
I am silly. I got 6.36 J for part 1. Now part 2 is a bit more confusing
 

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