Answer:
[tex]W_n_e_t=7.648512 \approx 7.6J[/tex]
[tex]K.E=0.8J[/tex]
[tex]v=0.7844645406 \approx 0.78m/s[/tex]
Explanation:
From the question we are told that
Mass of pitcher [tex]M= 2.6kg[/tex]
Force on pitcher [tex]f=8.8N[/tex]
Distance traveled [tex]48cm=>0.48m[/tex]
Coefficient of friction [tex]\mu=0.28[/tex]
a)Generally frictional force is mathematically given by
[tex]F=\mu N[/tex]
[tex]F=0.28*2.6*9.8[/tex]
[tex]F=7.1344N[/tex]
Generally work done on the pitcher is mathematically given as
[tex]W_n_e_t=W_f+W_F[/tex]
[tex]W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N[/tex]
[tex]W_n_e_t=4.224-3.424512[/tex]
[tex]W_n_e_t=0.799488\approx 0.8J[/tex]
b)Generally K.E can be given mathematically as
[tex]K.E= W_n_e_t[/tex]
Therefore
[tex]K.E=0.8J[/tex]
c)Generally the equation for kinetic energy is mathematically represented by
[tex]K.E=1/2mv^2[/tex]
[tex]0.8=1/2mv^2[/tex]
Velocity as subject
[tex]v=\sqrt{\frac{K.E*2}{m} }[/tex]
[tex]v=\sqrt{\frac{0.8*2}{2.6} }[/tex]
[tex]v=0.7844645406 \approx 0.78m/s[/tex]