Question

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

Answer 1

Answer:

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

       $K_{f}$ = 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

     $K_{f}$ = ½ m v2

     $K_{f}$ = ½ m (v₀² + 2 (F / m) d)

     $K_{f}$ = K₀  + F d

f)   v = √ 2$K_{f}$ / m