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Work and the Dot Product
A variable 1D force acts on an object of mass 2 kg, which is initially moving at 5 m/s to the right (along the positive x direction). The net force is given by:
F x = 20x2-10 i Newtons x
The force acts on the object as it displaced from x = 1 m to x = 4 m .
a) Findthespeedoftheobjectatx=4m.
b) Is there a gain or loss in kinetic energy or no loss in kinetic energy in the
displacement of the object? Explain.

Respuesta :

Answer:

a)  v_f = 5,06 m/s, b)  GAIN in kinetic energy.

Explanation:

a) For this exercise we will use the relationship between work and kinetic energy

           W = ΔK

Work is defined by

           W = F. d

bold indicates vectors

the displacement is

            d = x_f - x₀

            d = 4 -1

            d = 3i m

we calculate

          W = 20 10⁻² 3 i.i

let's remember that

         i.i = j.j = 1

         i.j = 0

           

          W = 6.0 10⁻¹ J

we substitute in the first equation

          W = K_f - K₀

          W = ½ m (v_f ² -v₀²)

          v_f ² = [tex]\frac{2W}{m} + v_o^2[/tex]

           

let's calculate

          v_f ² = 2 6.0 10⁻¹ /2 + 5²

          v_f = √25.6

          v_f = 5.06 m / s

             

b) we can see that the speed at the end of the movement is greater than the initial speed, therefore there is a GAIN in kinetic energy.