Answered

need help on these 2 an explanation of what formula’s to use would also be appreciated

need help on these 2 an explanation of what formula’s to use would also be appreciated class=

Answer :

angrykai23

Answer:

17. .343 seconds: 18.a .357: 18.b 64.5N

Explanation:

17. Since there is no friction, we can apply Newton's equation:

  • [tex]f=ma[/tex]
  • [tex]85N=5kg*a[/tex]
  • [tex]17 m/s^2=a[/tex]

We can then use the kinematic equation:

  • [tex]d=v_{i} *t+\frac{1}{2} at^2[/tex][tex]25N=U_{k} *70N[/tex]
  • [tex]1m=0+\frac{1}{2}*17m/s^2*t^2[/tex]

Solve for t:

  • [tex]{\frac{1}{8.5}s^2} =t^2[/tex]
  • [tex]\sqrt{\frac{1}{8.5} }=t=.343 seconds[/tex]

18.a We can use the friction formula:

  • [tex]F_{k} =U_{k} *F_{N}[/tex]
  • [tex]25N=U_{k} *70N[/tex]

Solving for kinetic friction of block mass 7kg:

  • [tex].357=U_{k}[/tex]

18.b We should add all forces that will counteract the rightward acceleration.

  • [tex]F_{a} =F_{x} -F_{N}[/tex]
  • [tex]Fr_{7kg} =25N : Fr_{2.5kg} =15.75N[/tex]

Use Newton's equation:

  • [tex]F=ma[/tex]
  • [tex]F_{a} =9.5kg*2.5m/s^2[/tex]
  • [tex]F_{a} =23.75N[/tex]

Solve for [tex]F_{x}[/tex]

[tex]F_{a} +F_{fr} =F_{x}[/tex]

[tex]23.75N+40.75N=F_{x} =64.5N[/tex]

Other Questions