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A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m. Neglecting friction between truck and road, determine the following:

a. The speed v.
b. The horizontal force exerted on the truck.

Respuesta :

cjrodriguez89

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        [tex]\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J[/tex]

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        [tex]v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s[/tex]

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        [tex]W = F*d[/tex]

  • We can solve for F, as follows:

        [tex]F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N[/tex]

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