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.]

m_1=5kg

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

Therefore

b

Generally the equation for the Tension  is mathematically given as

m=10kg

For more information on this visit

link

(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

it has to be b, about 97 percent.

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray