# Answer to Question #23213 in Mechanics | Relativity for Christian

Question #23213

What is the mass of block X?

Block A of mass 8.0 kg and block X are attached to a rope that passes over a pulley. A 50-N force P is applied horizontally to block A, keeping it in contact with a rough vertical face. The coefficients of static and kinetic friction between the wall and block A are μs = 0.40 and μk = 0.30. The pulley is light and frictionless. In the figure, the mass of block X is adjusted until block A descends at constant velocity of 4.75 cm/s when it is set into motion. What is the mass of block X?

A - 6.5 kg

B - 7.2 kg

C - 9.5 kg

D - 8.8 kg

E - 8.0 kg

THIS IS THE IMAGE: http://i102.photobucket.com/albums/m90/CMParts/f1q111g1_zpsf92b2525.jpg

Block A of mass 8.0 kg and block X are attached to a rope that passes over a pulley. A 50-N force P is applied horizontally to block A, keeping it in contact with a rough vertical face. The coefficients of static and kinetic friction between the wall and block A are μs = 0.40 and μk = 0.30. The pulley is light and frictionless. In the figure, the mass of block X is adjusted until block A descends at constant velocity of 4.75 cm/s when it is set into motion. What is the mass of block X?

A - 6.5 kg

B - 7.2 kg

C - 9.5 kg

D - 8.8 kg

E - 8.0 kg

THIS IS THE IMAGE: http://i102.photobucket.com/albums/m90/CMParts/f1q111g1_zpsf92b2525.jpg

Expert's answer

The velocity is constant, thusthe net force is zero.

mA * g - T - μk*P = 0

T = mX * g;

mX = (mA*g - μk*P) / g = (8 kg * 9.8 m/s^2 - 0.3 * 50 N) / (9.8 m/s^2) = 6.469

kg.

It corresponds approximately to 6.5 kg.

Answer: A

mA * g - T - μk*P = 0

T = mX * g;

mX = (mA*g - μk*P) / g = (8 kg * 9.8 m/s^2 - 0.3 * 50 N) / (9.8 m/s^2) = 6.469

kg.

It corresponds approximately to 6.5 kg.

Answer: A

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