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Block A of mass 5.0 kg and block X are attached to a rope which passes over a pulley, as shown in the figure. An 80-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 are μs = 0.40 and μk = 0.30. The pulley is light and frictionless. The mass of block X is adjusted until block A moves upward with an acceleration of 1.6 m/s2. What is the mass of block X?

Sagot :

MathPhys

Answer:

9.9 kg

Explanation:

According to Newton's second law of motion, the net force (ΣF) on an object is equal to the mass (m) times the acceleration (a). By drawing a free body diagram for each block, we can show all the forces acting on that block.

There are two forces on block X:

  • Weight force Mg pulling down, where M is the mass of block X
  • Tension force T pulling up

There are five forces on block A:

  • Weight force mg pulling down, where m is the mass of block A
  • Tension force T pulling up
  • Applied force P pushing left
  • Normal force N pushing right
  • Friction force Nμ pushing down, where μ is the coefficient of friction

Block X accelerates downwards. Sum of forces in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of forces on block A in the x direction:

∑F = ma

N − P = 0

N = P

Block A accelerates upwards. Sum of forces in the +y direction:

∑F = ma

T − Nμ − mg = ma

Substitute and solve for mass of block X, M:

Mg − Ma − Pμ − mg = ma

Mg − Ma − Pμ = mg + ma

M (g − a) − Pμ = m (g + a)

M (g − a) = m (g + a) + Pμ

M = [ m (g + a) + Pμ ] / (g − a)

Plug in values. Since block A is moving, use the kinetic friction coefficient.

M = [ 5.0 kg (9.8 m/s² + 1.6 m/s²) + 80 N × 0.30 ] / (9.8 m/s² − 1.6 m/s²)

M = 9.9 kg

View image MathPhys
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