Question

2.00 m/s2 T1 15.0 kg m1 T2 m2 20.0 kg 37.0°

          2.00 m/s2
T1
15.0 kg
m1
T2
m2 20.0 kg
37.0°

2.00 m/s2
T1
15.0 kg
m1
T2
m2 20.0 kg
37.0°

Added by Kyle A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prabhu Ramji

11/16/2022

Step 1

For the 15 kg block, the force due to gravity is F1 = m1*g = 15 kg * 9.8 m/s^2 = 147 N. For the 20 kg block, the force due to gravity is F2 = m2*g = 20 kg * 9.8 m/s^2 = 196 N. Show more…

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Two blocks, as shown in the figure, are connected by a cord of negligible mass that passes over a pulley with radius 0.250 m and moment of inertia I. The block on the frictionless inclined plane moves upward with constant acceleration of 2.00 m / s2. a) Determine T1 and T2, the tensions in the two parts of the rope. b) Find the moment of inertia of the pulley

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