Question

A wheel (radius = 0.25 m) is mounted on a frictionless, horizontal axis. The moment of inertia of the wheel about the axis is 0.040 kg.m . A light cord wrapped around the wheel supports a 0.50-kg object. The object is released from rest. What is the magnitude of the acceleration of the 0.50-kg object?

Answer 1

The magnitude of the acceleration of the 0.50-kg object is 9.8 m/s^2.

Step-by-step explanation:

The magnitude of the acceleration of the 0.50-kg object can be calculated using the equation:

a = F/m

Where F is the net force acting on the object and m is the mass of the object.

Since the object is released from rest, the net force is equal to the weight of the object:

F = mg

Substituting the values and solving for a:

a = (0.50 kg) * (9.8 m/s^2) / (0.50 kg) = 9.8 m/s^2

Therefore, the magnitude of the acceleration of the 0.50-kg object is 9.8 m/s^2.