Respuesta :
Explanation:
Formula for work done by sphere is as follows.
    [tex]W_{s} = \frac{1}{2} kx^{2}[/tex]
          = [tex]\frac{1}{2} 220 N/m \times (0.31 m)^{2}[/tex]
          = 10.57 J
Formula for work by friction is as follows.
     [tex]w_{f} = -fS[/tex]
          = [tex]-(\mu mg Cos \theta) S[/tex]
          = [tex]-(0.8 \times 5 kg \times 9.8 m/s^{2} \times Cos (18^{o})) \times 0.8 m[/tex]
          = [tex]-(0.8 \times 5 kg \times 9.8 m/s^{2} \times 0.951 \times 0.8 m[/tex]
          = -29.82 J
Work by gravity is given as follows.
     [tex]w_{g}[/tex] = mgh
           = [tex]5 kg \times 9.8 m/s^{2} \times 0.31 m[/tex]
           = 15.19 J
Hence, net work done will be given as follows.
     [tex]w_{s} + w_{f} + w_{g}[/tex]
    = 10.57 J + (-29.82) J + 15.19 J
    = -4.06 J
According to work energy theorem,
        w = [tex]\Delta k = \frac{1}{2}mv^{2}_{f}[/tex]
     -4.06 J = [tex]\frac{1}{2} \times 5 kg \times v^{2}_{f}[/tex]
       [tex]v_{f}[/tex] = 1.27 m/s
Thus, we can conclude that final velocity of the given mass is 1.27 m/s.
