Question

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a rope to a block with mass m2 = 0.825 kg as shown in the figure below. The rope goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is μk = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the rope does not stretch and does not slip on the pulley, and that the mass of the rope is negligible. (a). Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown (b). What is the angular speed of the pulley (in rad/s) after the block has moved this distance?

          A hanging weight, with a mass of m1 = 0.365 kg, is attached by a
rope to a block with mass m2 = 0.825 kg as shown in the figure
below. The rope goes over a pulley with a mass of M = 0.350 kg. The
pulley can be modeled as a hollow cylinder with an inner radius of
R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of
the spokes is negligible. As the weight falls, the block slides on
the table, and the coefficient of kinetic friction between the
block and the table is μk = 0.250. At the instant shown, the block
is moving with a velocity of vi = 0.820 m/s toward the pulley.
Assume that the pulley is free to spin without friction, that the
rope does not stretch and does not slip on the pulley, and that the
mass of the rope is negligible.
(a). Using energy methods, find the speed of the block (in m/s)
after it has moved a distance of 0.700 m away from the initial
position shown
(b). What is the angular speed of the pulley (in rad/s) after
the block has moved this distance?

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Added by Sara D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Baskar P

04/23/2022

Step 1

First, we need to calculate the total initial energy of the system. The initial energy consists of the kinetic energy of the block and the potential energy of the hanging weight. The kinetic energy of the block is given by (1/2)*m2*vi^2 and the potential energy of Show more…

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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a rope to a block with mass m2 = 0.825 kg as shown in the figure below. The rope goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is μk = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the rope does not stretch and does not slip on the pulley, and that the mass of the rope is negligible. (a). Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown (b). What is the angular speed of the pulley (in rad/s) after the block has moved this distance?

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Transcript

00:01 Hi, in this question the mass of the block 1 is given as 0 .365 kg.

00:10 M1 is given as 0 .365 kg and the mass of the second block m2 is given as 0 .825 kg.

00:27 The mass of the pulley, let us say capital m is given as 0 .35 kg.

00:40 The radius r1 is given as 0 .02 meters and r2 is given as 0 .03 meters.

00:58 Then the coefficient of kinetic friction mu k is given as 0 .25 and the initial velocity the i is given as 0 .820 meters per second the distance moved by the block 2 is given as 0 .820 meters per second the distance moved by the block 2 is given as 0 .7 meters we need to determine the final speed of the block 2.

01:39 Also we need to determine the final angular velocity of the pulley.

01:44 Now let us find out the mass moment of inertia or rotational inertia of the pulley.

01:51 That is, i is equal to 1 by 2 mass of the pulley multiply with r1 square plus r2 square.

02:07 R1 square plus r2 square.

02:12 Now plug the given values and determine the mass momentum of inertia.

02:18 1 by 2 mass of the pull is 0 .35.

02:23 R1 is 0 .02 square plus 0 .03 square.

02:33 On calculation we have taken the mass momentum of inertia i is equal to 2 .275 multiply with 10 to the power minus 4 kg meters square.

02:55 Now let us apply the work energy theorem and find the final velocity of the block 2.

03:04 According to the work energy theorem, the work done by the gravity plus the work done by the frictional force, wg plus wf must be equal to the change in kinetic energy of the system.

03:30 Change in kinetic energy of the entire system...

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