In Example 8-10, mass m1 = 2.40 kg is connected to mass m2 = 1.80 kg as shown. S

Question

In Example 8-10, mass m1 = 2.40 kg is connected to mass m2 = 1.80 kg as shown. Suppose the two masses start from rest and are moving with a speed of 1.85 m/s just before m2 hits the floor.

http://www.webassign.net/walker/08-10ex.gif


(a) If the coefficient of kinetic friction is µk = 0.360, what is the distance of travel, d, for the masses?


(b) How much conservative work was done on this system?



(c) How much nonconservative work was done on this system?

Explanation / Answer

a)now from the work energy principle work done by friction = change in potential energy of the system + change in kinetic energy of the system => m2*g*d = µk*m1*g*d + 1/2* (m1+m2)*v^2 > solving for d we have d = 0.7827 m = 78.27 cm

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