An atwood machine consists of two masses, ma = 7.0 kg and mb = 8.0 kg , connected by a cord that passes over a pulley free to rotate about a fixed axis. the pulley is a solid cylinder of radius r0 = 0.40 m and mass 0.70 kg .

Explanation:

Given Information

Block A mass, mA = 7 Kg

Block B mass, mB=8 Kg

Pulley mass, M=0.7 kg

Pulley Radius, R=0.4 m

I=moment of inertia=MR^2/2 =0.35R^2

Looking for

Acceleration of both blocks.

Percentage error of the two cases (the pulley’s moment of inertia is ignored and the pulley’s moment of inertia is not ignored).

1) mass acceleration a = g(mB – mA)/(mB + mA + I/R^2) = 9.8*1/(8+7+0.35) = 0.638 (m/s2), cylinder acceleration ε = a/R = 1.6m(s-2)

2)

In case the moment of inertia of the pulley is ignored mass acceleration a1 = 9.8*1/(8+7) = 0.653 (m/s2), % error = 100(aignored-a)/aignored = 2.30%.

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-lavira: )

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.