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A 3.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 28.0. with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s^2. If the coefficient of kinetic friction between

the block and the wall is 0.40, find

a) the work is done by the force on the block.

Answer in units of J.

b) the work done by gravity on the block.

b) the work done by gravity on the block.

A 30 kg block is pushed 20 m at a constant velocity up a vertical wall by a constant force applied at an angle of 280 with the horizontal as shown in the figure class=

Respuesta :

citlali18

Answer:

a)  [tex]w_{Fy} = 237.61713[/tex]J

b)  [tex]w_{g}=-58.86[/tex] J

c)  [tex]N=223.44641[/tex]N

Explanation:

Key: theta = θ kinetic friction = μ α = 180°

a) Work of Force put on block

Remember that you need both sides x & y

Equations:

  1. [tex]w_{F}=F_{x} cos[/tex]  |  [tex]w_{F}=F_{y} sin[/tex]θ
  2. [tex]_{net} F_{x} =0[/tex]      |  [tex]_{net} F_{y} =0[/tex]
  3. [tex]N=Fcos[/tex]θ   | [tex]F_{y} sin[/tex]μ[tex]N-mg=0[/tex]
  4.                     |  [tex]F_{y} sin[/tex][tex]F_{y} =\frac{mgd sin(theta)}{sin(theta)-(kinetic friction)cos(theta)}[/tex]-μ[tex]Fcos[/tex]θ[tex]=mg[/tex]
  5.                     |[tex]w_{Fy} =Fdsin[/tex]θ
  6.                     |  [tex]w_{Fy} =\frac{(3*9.81)sin(28)}{sin(28)-(0.40)cos(28)}[/tex]
  7.                     | [tex]w_{Fy} = 237.61713[/tex]J

b) Work of Gravity on block

  1. [tex]w_{g} =F_{g}dcos[/tex]α
  2. [tex]w_{g} =mgdcos[/tex]α
  3. [tex]w_{g}=(3)(9.81)(2)cos(180)[/tex]
  4. [tex]w_{g}=-58.86[/tex]J

c) Magnitud of Normal Force

  1. [tex]F=\frac{mg}{sin(theta)-(kinetic friction)cos(theta)}[/tex]
  2. [tex]F=\frac{3*9.81}{sin(28)-(0.40)cos(28)}[/tex]
  3. [tex]F = 253.06871[/tex]
  4. [tex]N=Fcos[/tex]θ
  5. [tex]N=(253.06871)cos(28)[/tex]
  6. [tex]N=223.44641[/tex]N

Hope it helps

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