Respuesta :

  • Mass=1167kg
  • Initial velocity=u=10m/s
  • Acceleration=a=4m/s^2
  • Work done=105J=W
  • Final velocity=v=?
  • Force=F
  • Distance=d

Apply Newton's second law

[tex]\\ \tt\hookrightarrow F=ma[/tex]

[tex]\\ \tt\hookrightarrow F=1167(4)=4668N[/tex]

Now

[tex]\\ \tt\hookrightarrow W=Fd[/tex]

[tex]\\ \tt\hookrightarrow d=\dfrac{W}{F}[/tex]

[tex]\\ \tt\hookrightarrow d=\dfrac{105}{4668}[/tex]

[tex]\\ \tt\hookrightarrow d=0.022m[/tex]

Now

  • d be s

According to third equation of kinematics

[tex]\\ \tt\hookrightarrow v^2=u^2+2as[/tex]

[tex]\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)[/tex]

[tex]\\ \tt\hookrightarrow v^2=100+8(0.022)[/tex]

[tex]\\ \tt\hookrightarrow v^2=100+0.176[/tex]

[tex]\\ \tt\hookrightarrow v^2=100.176[/tex]

[tex]\\ \tt\hookrightarrow v=10.001m/s[/tex]

Otras preguntas