Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.098

Determine the acceleration of the system

Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a = the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer: 0.665 m/s² (nearest thousandth)

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.098 Determine the acceleration of the system

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Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.098

Determine the acceleration of the system