What is Work-energy theorem: Definition and 95 Discussions

In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force.
For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force F is constant and the angle between the force and the displacement s is θ, then the work done is given by:




W
=
F
s
cos
⁡

θ



{\displaystyle W=Fs\cos {\theta }}
Work is a scalar quantity, so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J), the same unit as for energy.

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  1. P

    Misunderstanding Work-energy theorem and center of mass properties

    To do this apparently, you need to use the work-energy theorem. You can calculate work done by gravity easily. However it was said that work done by the reaction forces from the hinge is zero, I don't get why. Reaction Force from the hinge is an external force on the rod, and all external...
  2. P

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  3. Argonaut

    Work-Energy Theorem and Friction

    My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt: Let block 1 be the vertically moving block and let block 2 be the horizontally moving one. Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0...
  4. Argonaut

    Understanding Work-Energy Theorem: Solving an Exercise with Different Solutions

    My solution is different from the official solution and I don't understand what I did wrong. Here is my solution: The magnitude of the initial velocity is ##|v_0| = 12.0~\rm{m/s}##, so the vertical component of the initial velocity is ##v_{0-y} = (12.0 \sin{25^{\circ}})~\rm{m/s}##. Then I use...
  5. Philip551

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  6. C

    I The far reaching ramifications of the work-energy theorem

    The work-energy theorem is the connection between expressing mechanics taking place in terms of force-and-acceleration, ##F=ma## and representing mechanics taking place in terms of interconversion of kinetic energy and potential energy. The following statements are for the case that there is a...
  7. sachin

    Applying the Work-Energy theorem to a system

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  8. Z

    Possible to use work-energy theorem from a non-inertial frame?

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  9. P

    What exactly does the Work-Energy theorem say?

    My research leads to 2 slightly different equations. See equations 1 and 2 attached. Also, for equation 1 should Wext be Wnet ext ?
  10. Soyuz42

    Work Done by Elevator Cable in Sample Problem 7-6

    As stated, part (a) says that the work done by the gravitational force ##\vec{F_g}## is 59 kJ. If ##W_T## is the work done by the elevator cable during the 12 m fall, then using the work-kinetic energy theorem, \begin{align*} K_f -K_i &= W_g + W_T\\ \frac12m({v_f}^2 - {v_i}^2) &= 59000 + W_T\\...
  11. ArtVandolay

    I Relativistic Work-Energy Theorem: Deriving Griffiths' Formula

    In deriving the work-energy theorem, Griffiths does the following: ##\frac{d\mathbf{p}}{dt}\cdot\mathbf{u} = \frac{d}{dt}\bigg(\frac{m\mathbf{u}}{\sqrt{1-u^2/c^2}}\bigg)\cdot\mathbf{u}=\frac{m\mathbf{u}}{(1-u^2/c^2)^{3/2}}\cdot\frac{d\mathbf{u}}{dt}## I may have forgotten something essential...
  12. n3pix

    What does this mean ##U=\int F\times d\vec{r}## in the Work-Energy Theorem?

    Hello, I'm newly discovering the world of the Energy. My question is about the equation ##U=\int \vec{F}\times d\vec{r}=-\int \vec{F}_{s}\times d\vec{r}##. Can you tell me what does this equation means? Thanks!
  13. A

    Is this work-energy theorem for a system correct?

    Homework Statement [/B] A 4 kg mass slides 2 m over a horizontal surface with force of kinetic friction of 2N, initial velocity is 5 m/s. Find it's final velocity. My problem... we've been asked to solve this problem using the Work-Energy theorem, and we've been given it in the form of: W =...
  14. G

    When is Gravitational Potential Energy Considered in the Work-Energy Theorem?

    When using the work-energy theorem (Wnet=ΔE), when do you take gravitational potential energy into account? Change in energy implies all types of energy involved, but in what cases would PEg be a part of it?
  15. vdvdlk

    Work-Energy theorem if the particle´s mass depends on time

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  16. donaldparida

    Generalized version of work-energy theorem

    I know that for rigid bodies only the work-energy theorem states that the net work done on the body equals the change in kinetic energy of the body since a rigid body has no internal degrees of freedom and hence no other forms of energy such as potential energy. Is there a most generalized form...
  17. D

    Question About Work-Energy Theorem

    Very simple question. So I am on a homework problem, and I want to make sure that I am using this theorem correctly. My book states that the Work-Kinetic Energy Theorem is: W=ΔK Now the solution to this problem involved multiple forces and thus each force is doing work. So my question is, is...
  18. S

    Doubts on Work-Energy theorem for a system

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  19. normal_force

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  20. D

    Use the work-energy theorem to derive an expression for v^2

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  21. J

    Work-Energy Theorem problem

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  22. D

    Why is the work done and force applied different in the Work-Energy Theorem?

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  23. F

    Why do I not get the same result when I use change in PE?

    I found the answer to this problem using the change in KE, but when I try to relate the work done on the 12.0-N block in terms of potential energy i don't get the same result. Is the change potential energy not equal to the work done? I would also like to know what situations is the change PE...
  24. S

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  25. H

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  27. A

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  28. DiracPool

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  29. I

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    Wnet = ΔKE By this equation, if I lift a 1 kilogram book at rest from the ground and place it to be at rest on a table 10 meters above the ground, no net work has been done on the book. (Its kinetic energy before and after is zero.) However, its potential energy has changed by mgh or 1kg *...
  30. C

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  31. DiracPool

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  32. S

    Why Is Mass Halved in the Work-Energy Theorem Derivation?

    Hello, Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?: ##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##
  33. K

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  34. S

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  35. D

    Minimum work and work-energy theorem

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  36. skate_nerd

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  37. S

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  38. S

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  39. M

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  40. Y

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  41. A

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  42. Vorde

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  43. L

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  44. P

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    Help! I am a starter physicist and I need hlep on this question stated thus: Homework Statement A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the +x axis. As a result...
  45. jaketodd

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  46. E

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  47. J

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  48. J

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  49. T

    Work-Energy Theorem question with not much detail to work with

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  50. P

    Work-Energy Theorem: Understanding Vertical Movement

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