A 0.414 kg object is at rest at the origin of a coordinate system. A 2.08 N force in the positive x direction acts on the object for 1.57 s. What is the velocity at the end of this interval

Respuesta :

Answer:

The value is [tex]v = 7.89 \ m/s[/tex]

Explanation:

From the question we are told that

  The mass  of the object is [tex]m = 0.414 \ kg[/tex]

   The force is [tex]F = 2.08 \ N[/tex]

  The time taken is  [tex]t = 1.57 \ s[/tex]

Generally the force acting on the object according to Newton's second law is

        [tex]F = \frac{ m (v -u )}{t}[/tex]

=>     [tex]v = \frac{F * t}{m} + u[/tex]

Here u is the initial velocity with value  0 m/s since the object was initially at rest

        [tex]v = \frac{2.08 * .57}{0.414 } + 0[/tex]

     [tex]v = 7.89 \ m/s[/tex]