A 28.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system begins to move.(adsbygoogle = window.adsbygoogle || []).push({});

a) Calculate the mass of sand added to the bucket

b) Calculate the acceleration for the system.

a)

28.0 kg block weighs (28.0 kg)(9.80 m/s2) = 274.4 N

Ffr(max) = µsFN

Ffr(max) = (.450)(274.4 N) = 123.48 N

F = ma

123.48 N = m(9.80 N/kg)=12.6 kg

12.6 kg – 1.0 kg=11.6 kg of sand

b)

Force of Kinetic Friction between Block and Table

Ffr = µkFN

Ffr = (.320)(274.4 N) = 87.808 N

Force of Bucket

F=ma

123.48 - T= 12.6a

Force of Block

F = ma

T - 87.808 N= 28a

Substituting First Expression into Second Expression and Solve for T:

123.48 - T = 12.6a, T = 123.48 - 12.6a

T - 87.808 N = 28a

(123.48 - 12.6a) - 87.808 N = 28a

35.672 = 40.6a

a = 0.879 m/s2

Does this all look correct?

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# Newtons Laws- Double checking

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