A block of mass 0.50 kg is on a table and attached to a block of mass 0.25 kg by

Question

A block of mass 0.50 kg is on a table and attached to a block of mass 0.25 kg by a taut, massless string. The 0.25-kg block hangs by means of the string, which goes over a pulley. The 0.25-kg block is released, and it descends from rest at a constant acceleration, travelling 1.00 m in 1.25 s. What is the coefficient of kinetic friction between the table and the 0.50-kg block? A block of mass 0.50 kg is on a table and attached to a block of mass 0.25 kg by a taut, massless string. The 0.25-kg block hangs by means of the string, which goes over a pulley. The 0.25-kg block is released, and it descends from rest at a constant acceleration, travelling 1.00 m in 1.25 s. What is the coefficient of kinetic friction between the table and the 0.50-kg block?

Explanation / Answer

s = 0.5at^2

so acceleration a = 1.28m/s^2

0.25g - T = 0.25a

T - u0.5*g = 0.5a

adding both

0.25g - u0.5g = 0.75a

so u = 0.3

so coefficient of kinetic friction between table and block = 0.3

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