10.(20 points) A block of mass m1 23.9 kg is at rest on a plane inclined at 30.0

Question

10.(20 points) A block of mass m1 23.9 kg is at rest on a plane inclined at 30.0° above the horizontal. The block is connected via a rope and mass less pulley system to another block of mass m2-25.1kg, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane is k 0.086. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.50 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

m1 = 23.9 kg

m2 = 25.1 kg

uk = 0.086

t = 1.5 s

When the system is moving the net (unbalanced) force acting (Fr) is ..
Fr = (Weight of m2) - (weight component of m1 down incline) - (kinetic friction)
Fr = (m2*g) - (m1*g*sin30) - (k*m1*g*cos30)
Fr = (25.1*9.8) - (23.9*9.8*sin30) - (0.086*23.9*9.8*cos30)

Fr = 111.4 N

Acceleration of joined mass, a = Fr / (m1 + m2)
a = 111.4/(23.9 + 25.1) = 2.27 m/s²

Distance moved .. d = ut + ½.at²
d = 0 + (½ *2.27*1.5²) = 2.55 m

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