Question

A block of mass m2 = 15 kg on a rough 30°-inclined plane is connected to a 5-kg mass (m1) by a string of negligible mass passing over a pulley that is shaped like a disk. The 2-kg pulley has radius 15 cm and rotates about its symmetry axis of rotation. The string does not slip on the pulley and causes the pulley to rotate about a fixed horizontal axle through its center of mass. When this system is released from rest, the block (m2) moves at a uniform linear acceleration. The coefficient of kinetic friction between the 15-kg block and the incline surface is 0.10. (a) Find the linear acceleration of the system. (b) Calculate the tension T1, in the string supporting m1, and tension T2, in the string that supports m2. (c) Use energy methods to find the speed of the blocks when the 5-kg block has moved through a vertical displacement of 3 m from rest. (a) [a = 0.56 m/s2]; (b) [T1 = 52.4 N; T2 = 51.8 N]; (c) [v = 1.83 m/s]

Answer 1

The situation can be visualized as

From the figure

The equation of motion for the block 1 is

The equation of motion for the block 2 is

The moment of inertia of the pully is

The net torque on the disk is

The equation of motion of the pully is

Putting (3) in (2)

Adding (1) to it

PART A:

Putting the values in the equation

So, the acceleration is 0.56 m/s².

PART B:

We have

And we have

PART C:

The force acting on block 2 is

Work done by the force in moving the block by 3 m is

The energy is converted to kinetic energy. So,