Answer with Explanation:
We are given that
Mass of crate,m=9.6 kg
Speed,u=1.46 m/s
Force,F=102 N
[tex]\theta=19.9^{\circ}[/tex]
[tex]\mu=0.4[/tex]
Distance,d=5.02 m
a.Work done=[tex]Fd=mgsin\theta d[/tex]
Where [tex]g=9.81 m/s^2[/tex]
[tex]W=9.6\times 9.81sin19.9\times 5.02=160.7 J[/tex]
b.Increase in internal energy of the crate,E=[tex]\mu mg dcos\theta[/tex]
[tex]E=0.4\times 9.81\times 5.02\times 9.6cos19.9[/tex]
[tex]E=177.75 J[/tex]
c.Work done by 102 N=[tex]Fd=102\times 5.02=512.04 J[/tex]
d.Change in kinetic energy=-160.7+512.04-177.75=173.6 J
e.[tex]\frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K.E[/tex]
[tex]\frac{1}{2}m(v^2-u^2)=\Delta K.E[/tex]
[tex]v^2-u^2=\frac{2\Delta K.E}{m}[/tex]
[tex]v^2=\frac{2\Delta K.E}{m}[/tex][tex]+u^2[/tex]
[tex]v=\sqrt{\frac{2\Delta K.E}{m}+u^2}[/tex]
[tex]v=\sqrt{\frac{2\times 173.6}{9.6}+(1.46)^2}=6.2 m/s[/tex]
