Answer:
a. 7.38 N b. 40.87 N c. 0.113 kg-m²
Explanation:
a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest), s = distance moved = 1.30 m and t = time = 0.850 s.
Substituting these values into s,
1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a
1.30 = 0.36125a
a = 1.30/0.36125 = 3.6 m/s²
Substituting this into T, we have
T = ma = 2.05 kg × 3.6 m/s² = 7.38 N
b. Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is
T - mg = ma
T = m(a + g)
substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg
T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N
c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.
α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²
Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m
From the equation above, I = Tr/α
Substituting the variables we have
I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²
