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The 2-kg package leaves the conveyor belt at with a speed of vA = 1 m/s and slides down the smooth ramp. Determine the required speed of the conveyor belt so that the package can be delivered without slipping on the belt. Also, find the normal reaction the curved portion of the ramp exerts on the package.

Respuesta :

Answer:

Incomplete question: Find the normal reaction the curved portion of the ramp exerts on the package at B if pB = 2 m, the height is 4 m

Answer: The required speed is 8.9107 m/s

The normal reaction is 99.0006 N

Explanation:

Given data:

m = 2 kg

vA = speed = 1 m/s

h = 4 m

g = gravity = 9.8 m/s²

Questions: Determine the required speed of the conveyor, vc = ?

Find the normal reaction, Nc = ?

The required speed applying the conservation of energy:

[tex]\frac{1}{2} mv_{A}^{2} +mgh=\frac{1}{2} mv_{c} ^{2}[/tex]

Solving for vc:

[tex]v_{c} =\sqrt{v_{A}^{2}+2gh } =\sqrt{1^{2}+(2*9.8*4) } =8.9107m/s[/tex]

The normal reaction applying the equilibrium of forces:

[tex]N_{c} -mg=m(\frac{v_{c}^{2} }{p_{B} } )[/tex]

[tex]N_{c} =2*(\frac{8.9107^{2} }{2} )+(2*9.8)=99.0006N[/tex]

Cricetus

The required speed will be "8.9107 m/s" and normal reaction applying is "99.0006 N".

Given values are:

  • Mass, m = 2 kg
  • Speed, vA = 1 m/s
  • Height, h = 4 m
  • Gravity, g = 9.8 m/s²

By using the conservation of energy, we get

→ [tex]\frac{1}{2}mv_A^2+mgh = \frac{1}{2} mv_c^2[/tex]

then,

→ [tex]v_c = \sqrt{v_A^2+2gh}[/tex]

      [tex]= \sqrt{1^2+(2\times 9.8\times 4)}[/tex]

      [tex]= 8.9107 \ m/s[/tex]

hence,

The normal reaction applying:

→ [tex]N_c - mg = m(\frac{v_c^2}{pB} )[/tex]

            [tex]N_c = 2\times (\frac{8.9107^2}{2} )+ 2\times 9.8[/tex]

                 [tex]= 99.0006 \ N[/tex]

Thus the above approach is right.  

Learn more about speed here:

https://brainly.com/question/15683419

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