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a weight of mass 1.74 kg is suspended by a string wrapped around a pulley wheel, which consists of a solid disk of mass 4.22 kg and radius 1.30 m. the system is released from rest. over what vertical distance does the hanging mass move in 3.0 seconds? ignore friction and drag forces, and assume that the string does not slip.

User Adam Kis
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Final answer:

The hanging mass will move over a vertical distance of approximately 12.857 meters in 3.0 seconds.

Step-by-step explanation:

First, we need to calculate the moment of inertia of the pulley wheel. The moment of inertia of a solid disk is given by the formula I = (1/2) * m * r^2, where m is the mass of the disk and r is the radius.

Plugging in the values, we get I = (1/2) * 4.22 kg * (1.30 m)^2

= 3.4336 kg·m^2.

Next, we can calculate the acceleration of the system using Newton's second law, F = m * a.

The force that causes the acceleration is the tension in the string. The tension in the string is equal to the weight of the hanging mass, which is given by T = m * g, where m is the mass of the hanging mass and g is the acceleration due to gravity.

Plugging in the values, we get T = 1.74 kg * 9.8 m/s^2

= 17.052 kg·m/s^2.

Now we can calculate the acceleration of the system: 17.052 kg·m/s^2 = (4.22 kg + 1.74 kg) * a.

Solving for a, we get a = 17.052 kg·m/s^2 / 5.96 kg ≈ 2.8604 m/s^2.

Finally, we can calculate the distance the hanging mass moves in 3.0 seconds using the equation of motion, d = v0 * t + (1/2) * a * t^2, where v0 is the initial velocity of the hanging mass (which is 0 m/s because it starts from rest), a is the acceleration of the system, and t is the time.

Plugging in the values, we get d = 0 + (1/2) * 2.8604 m/s^2 * (3.0 s)^2

= 12.857 m.

User Librik
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