quennquerida
contestada

the coefficient of static friction between mass mA

and the table is 0.40, whereas the coefficient of kinetic friction

is 0.20.

(a) What minimum value of mA will keep the system from

starting to move?

(b) What value(s) of mA will keep the system moving at

constant speed?

[Ignore masses of the cord and the (frictionless) pulley.]​

the coefficient of static friction between mass mAand the table is 040 whereas the coefficient of kinetic friction is 020 a What minimum value of mA will keep t class=

Respuesta :

MathPhys

Answer:

(a) 5.0 kg

(b) 10 kg

Explanation:

Draw a free body diagram for each block.  There are 4 forces on block A:

Weight force mAg pulling down,

Normal force N pushing up,

Tension force T pulling right,

and friction force Nμ pushing left.

There are 2 forces on block B:

Weight force mBg pulling down,

and tension force T pulling up.

Whether the system is just starting to move, or moving at constant speed, the acceleration is 0.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = 0

mBg = T

Sum of forces on A in the +y direction:

∑F = ma

N − mAg = 0

N = mAg

Sum of forces on A in the +x direction:

∑F = ma

T − Nμ = 0

T = Nμ

Substitute:

mBg = mAg μ

mA = mB / μ

(a) When the system is just starting to move, μ = 0.40.

mA = 2.0 kg / 0.40

mA = 5.0 kg

(b) When the system is moving at constant speed, μ = 0.20.

mA = 2.0 kg / 0.20

mA = 10 kg

okpalawalter8
  • We have that minimum value of mA will keep the system from  starting to move is

m_1=5kg

  • The value(s) of mA will keep the system moving at  constant speed is

m=10kg

From the question we are told

the coefficient of static friction between mass mA  and the table is 0.40, where as the coefficient of kinetic friction  is 0.20.

a)  

Generally the equation for the Tension  is mathematically given as

T=mg

Where

[tex]m_1g=m_2g[/tex]

Therefore

[tex]m_1=\frac{2.0}{0.4}\\\\m_1=5kg[/tex]

b

Generally the equation for the Tension  is mathematically given as

[tex]T=f\\\\T=u_km_1g\\\\\m_1=\frac{m_2}{u}\\\\m_1=\frac{2}{0.2}[/tex]

m=10kg

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https://brainly.com/question/19694949

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