Horizontal force of 50N accelerates a block of mass 3.0kg up an incline with an A=2.0ms.The anglewith horizontal is 30 degree.a)force of friction between the box and the incline b)the coefficient of kinetic friction betweenbox and the inclinehelpexplain?

1 Answer
Nathan L.
Aug 7, 2017

(a) #f_k = 29.3# #"N"#

(b) #mu_k = 1.15#

Explanation:

We're asked to find

  • (a) the retarding friction force acting on the block as it is moved up the incline

  • (b) the coefficient of kinetic friction between the block and incline

  • (a)

I'll call the positive #x#-axis as directed up the incline, and the positive #y#-axis as perpendicular to the incline.

The horizontal forces acting on the block are

  • the applied force of #50# #"N"#, directed up the incline

  • the downward gravitational force, directed down the incline and equal to #mgsintheta#

  • the retarding friction force (#f_k#), also directed down the incline because it opposes motion

Our net horizontal force equation is thus

#ul(sumF_x = F_"applied" - mgsintheta - f_k = ma_x#

We're given

  • #m = 3.0# #"kg"#

  • #a_x = 2.0# #"m/s"^2#

So the net horizontal force acting on the object is

#sumF_x = ma_x = (3.0color(white)(l)"kg")(2.0color(white)(l)"m/s"^2) = color(red)(ul(6.0color(white)(l)"N"#

Thus, we have

#sumF_x = F_"applied" - mgsintheta - f_k = color(red)(6.0color(white)(l)"N"#

Plug in the known values below:

  • #F_"applied" = 50# #"N"#

  • #m = 3.0# #"kg"#

  • #g = 9.81# #"m/s"^2#

  • #theta = 30^"o"#:

#overbrace(50color(white)(l)"N")^(F_"applied") - overbrace((3.0color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(30^"o"))^(mgsintheta) - f_k = overbrace(color(red)(6.0color(white)(l)"N"))^(sumF_x)#

#50color(white)(l)"N" - 14.7color(white)(l)"N" - f_k = color(red)(6.0color(white)(l)"N"#

Therefore,

#color(blue)(ulbar(|stackrel(" ")(" "f_k = 29.3color(white)(l)"N"" ")|)#

#" "#

  • (b)

The equation relating the friction force and the coefficient of kinetic friction is

#ul(f_k = mu_kn#

where

  • #f_k = color(blue)(29.3color(white)(l)"N"#

  • #mu_k# is the coefficient of kinetic friction

  • #n# is the magnitude of the normal force exerted by the incline plane, equal to

#ul(n = mgcostheta#

So

#n = (3.0color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(30^"o") = color(green)(ul(25.5color(white)(l)"N"#

We can rearrange the above equation to solve for the coefficient:

#mu_k = (f_k)/n#

Plugging in known values:

#mu_k = (color(blue)(29.3color(white)(l)"N"))/(color(green)(25.5color(white)(l)"N")) = color(blue)(ulbar(|stackrel(" ")(" "1.15" ")|)#

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