A 350-g air track cart on a horizontal air track is attached to a string that goes over a pulley with a moment of inertia of 6.00 × 10^-6 kg∙m² and a radius of 1.35 cm. The string is pulled vertically downward by a force of 2.50 N. What is the tension in the string between the pulley and the cart? A) 4.58 N B) 2.50 N C) 1.85 N D) 2.29 N E) 1.74 N

Question

Quizplus · Accepted Answer

First, we need to find the moment of inertia of the pulley in SI units:
6.00 × 10^-16 kg∙m^2

Next, we need to find the torque on the pulley from the force applied by the string:
τ = rF = (0.0135 m)(2.50 N) = 0.03375 N∙m

Using the equation τ = Iα, we can find the angular acceleration of the pulley:
α = τ/I = 0.03375 N∙m/(6.00 × 10^-16 kg∙m^2) = 5.625 × 10^13 rad/s^2

Since the string is not slipping on the pulley, the tension in the string is equal to the force applied by the cart:
Tension = force applied by cart = mg = (0.350 kg)(9.81 m/s^2) = 3.4335 N

However, since the string is being pulled downward by the 2.50 N force, the net force on the cart is:
Fnet = Tension - 2.50 N

Using Newton's second law, we can find the acceleration of the cart:
Fnet = ma
Tension - 2.50 N = (0.350 kg)a
a = (Tension - 2.50 N)/(0.350 kg)

Now, we can use the kinematic equation:
vf^2 = vi^2 + 2ad

where vf = 0 (since the cart stops at the end of the track), vi = 0 (since the cart starts from rest), and d is the length of the air track (not given in the problem).

Solving for a, we get:
a = -vi^2/(2d) = -0/(2d) = 0

This means that the acceleration of the cart is zero, and therefore:
Tension - 2.50 N = 0
Tension = 2.50 N

Therefore, the correct answer is choice D.

A 350-G Air Track Cart on a Horizontal Air Track