Question

A block of mass m1=3.0kg rests on a frictionless horizontal surface. A second block of m2=2.0kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block . the blocks are released from rest . determine the displacement of the velocity of the first block 1.2 s after the release of the blocks, assuming the first block doesn't run out of the room on the table and the second block doesn't land on the floor?

A) 23.5m/s
B) 12m/s
C) 33.7m/s
D) 6.7m/s​

Answer 1

To solve this problem, we can use the principles of Newton's laws of motion and the conservation of energy.

At the moment of release, the second block will start to accelerate downwards due to gravity, and the first block will start to move to the right due to the tension in the rope. Since the surface is frictionless, there is no horizontal force acting on the first block once it starts moving.

Using the free-body diagrams for the two blocks, we can write the following equations of motion:

For the second block:

m2g - T = m2a

where g is the acceleration due to gravity, T is the tension in the rope, a is the acceleration of the second block, and m2 is the mass of the second block.

For the first block:

T = m1a

where m1 is the mass of the first block and a is its acceleration.

Since the two blocks are connected by a rope, they must have the same acceleration, so we can set the two equations for acceleration equal to each other:

m2g - T = m1a

T = m1a

m2g - m1a = T = m1a

Solving for a, we get:

a = (m2/m1 + m2)g

We can also use the conservation of energy to find the final velocity of the first block after 1.2 seconds. At the moment of release, the total mechanical energy of the system is given by:

E = m1gh

where h is the initial height of the second block. As the blocks move, the potential energy of the second block is converted into the kinetic energy of both blocks. At the end of the 1.2 seconds, all of the potential energy will be converted into kinetic energy, so we can write:

E = (1/2)m1v^2 + (1/2)m2v^2

where v is the final velocity of the first block.

Solving for v, we get:

v = sqrt(2gh(m1+m2)/m1)

Plugging in the given values, we get:

a = (2/5)g ≈ 3.92 m/s^2

v = sqrt(2gh(m1+m2)/m1) ≈ 2.36 m/s

Therefore, the displacement of the velocity of the first block 1.2 s after the release of the blocks is approximate:

vt + (1/2)at^2 = 2.361.2 + (1/2)3.92(1.2)^2 ≈ 5.52 m/s

So the answer is not given in the options.