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Review the work-energy theorem and apply it to a simple problem.

If you push a particle of mass M in the direction in which it is already moving, you expect the particle's speed to increase. If you push with a constant force F, then the particle will accelerate with acceleration a=F/M (from Newton's 2nd law).

a. Enter a one- or two-word answer that correctly completes the following statement.

If the constant force is applied for a fixed interval of time t, then the _____ of the particle will increase by an amount at.

b. Enter a one- or two-word answer that correctly completes the following statement.

If the constant force is applied over a given distance D, along the path of the particle, then the _____ of the particle will increase by FD.

c. If the initial kinetic energy of the particle is Ki, and its final kinetic energy is Kf, express Kf in terms of Ki and the work W done on the particle.

d.

In general, the work done by a force F⃗ is written as

W=∫fiF⃗ (r⃗ )⋅dr⃗ .

Now, consider whether the following statements are true or false:

The dot product assures that the integrand is always nonnegative.

The dot product indicates that only the component of the force perpendicular to the path contributes to the integral.

The dot product indicates that only the component of the force parallel to the path contributes to the integral.

Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t).

e. Assume that the particle has initial speed vi. Find its final kinetic energy Kf in terms of vi, M, F, and D.

f.

What is the final speed of the particle?

Express your answer in terms of Kf and M.

Respuesta :

Answer:

a) particle speed increases , b) work and energy will increase , c) [tex]K_{f}[/tex] = W -K₀ , e)  [tex]K_{f}[/tex] = K₀  + F d,  f)   v = √ 2[tex]K_{f}[/tex] / m

Explanation:

The Work and Energy Theorem stable a linear relationship between the two

               W = ΔK

a) particle speed increases

b) work and energy will increase

c) of the initial relationship

      W = [tex]K_{f}[/tex]- Ko

       [tex]K_{f}[/tex] = W -Ko

d)

False The scalar product can be positive, negative or zero

False. The perpendicular component gives a zero job, so it does not contribute to the integral

True. The parallel component gives a nonzero result

e) of Newton's second law

       a = F / m

Of kinematics

      v² = v₀² + 2 a d

       v²= v₀² + 2 F / m d

     [tex]K_{f}[/tex] = ½ m v2

     [tex]K_{f}[/tex] = ½ m (v₀² + 2 (F / m) d)

     [tex]K_{f}[/tex] = K₀  + F d

f)   v = √ 2[tex]K_{f}[/tex] / m

The work-energy theorem is reviewed for the partial which is under the push force in a direction, where it is already moving.

What is work energy theorem?

According to the work energy theorem, the sum of all the forces acting on a body to do a work is equal to the change in the kinetic energy of the body.

Lets, fill the blank first with an appropriate word, which completes the statement,

  • a) If the constant force is applied for a fixed interval of time t, then the velocity of the particle will increase by an amount at.
  • b) If the constant force is applied over a given distance D, along the path of the particle, then the work-done of the particle will increase by FD.
  • c) If the initial kinetic energy of the particle is Ki, and its final kinetic energy is Kf, express Kf in terms of Ki and the work W done on the particle.

It can be expressed as,

[tex]W=k_f-k_i\\k_f=W+k_i[/tex]

  • d) In general, the work done by a force F⃗ is written as

[tex]W=\int\limits^f_i {\vec F(\vec r)} \ . \vec dr[/tex]

Now, consider whether the following statements are true or false:

The dot product assures that the integrand is always nonnegative-This is a false statement.

The dot product indicates that only the component of the force perpendicular to the path contributes to the integral-This is a false statement.

The dot product indicates that only the component of the force parallel to the path contributes to the integral-This is a true or correct statement.

  • e. The final kinetic energy Kf in terms of vi, M, F, and D,

The initial kinetic energy is,

[tex]k_i=\dfrac{1}{2}Mv_i^2[/tex]

Thus, the final kinetic energy can be given as,

[tex]k_f=k_i+FD\\k_f=\dfrac{1}{2}Mv_i^2+FD[/tex]

  • f. The final speed of the particle-

Final kinetic energy can be given as,

[tex]K_f=\dfrac{1}{2}Mv_f^2\\v_f=\sqrt{\dfrac{2K_f}{M}}[/tex]

Hence, the work-energy theorem is reviewed for the partial which is under the push force in a direction, where it is already moving.

Learn more about the work energy theorem here;

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