kk We use the free-body diagrams to apply Newton\'s second law. For m1, T - f1 =
ID: 2114640 • Letter: K
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kk
We use the free-body diagrams to apply Newton's second law. For m1, T - f1 = m1a, and for m2, F - T - f2 = m2a. The definition of the coefficient of kinetic friction gives us the following equations. Remember that n1 = m1g and n2 = m2g. f1 = mu kn1 = mu km1g = ( )( kg)(9.80 m/s2) = N f2 = mu n2 = mu m2g = ( )( kg)(9.08 m/s2) = N If the coefficient of friction were zero, the acceleration would be approximately 2 m/s2. So we expect acceleration to be about 1 m/s2. The tension will be less than half of 72.0 N since m2 is greater than m1. Because the rope has constant length, both blocks move the same number of centimeters per second, so they move with the same acceleration. To find this acceleration and the tension, we treat the two blocks as separate accelerating objects. The free-body diagrams for m1 and m2 are shown below. The magnitude of the tension force exerted by m1 on m2 is equal to equal to the magnitude of the tension force exerted by m2 on m1. The magnitude of the normal force for m1 is less than less than the magnitude of the normal force for m2. The magnitude of the friction force for m2 is greater than greater than the magnitude of the friction force for m1.Explanation / Answer
T - f1 = m1a..........(1)
F - T - f2 = m2a.........(2).
If uk = 0
then f1 = f2 = 0
a = 2 m/s^2
Put in (1) and(2)
T = 2m1
F = 2( m1 + m2)
F = 72 N
(m1 + m2) = 36 Kg
If friction is present then a = 1 m/s^2
Adding eq 1 and 2
F - uk( m1 + m2)*g = (m1+ m2)*a
72 - 36uk*9.8 = 36
uk = 1/9.8 = 0.102
m1+ m2 = 36 kg
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