A block of mass m1 = 3.2 kg rests on a frictionless horizontal surface. A second block of mass m2 = 1.9 kg 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

a) Find the acceleration of the two blocks after they are released.

block 1:

magnitude

m/s2



direction

---Select--- to the right to the left

block 2:

magnitude

m/s2



direction

---Select--- upward downward

(b) What is the velocity of the first block 1.8 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor?

magnitude

m/s

direction

---Select--- to the right to the left

(c) How far has block 1 moved during the 1.8-s interval?
m

(d) What is the displacement of the blocks from their initial positions 0.35 s after they are released?

block 1:

magnitude

m

direction

---Select--- to the right to the left

block 2:

magnitude

m

direction

---Select--- up down

To find the answers to these questions, we can use Newton's laws of motion.

First, let's find the acceleration of the two blocks after they are released. We can start by considering the forces acting on each block.

For block 1 (m1 = 3.2 kg), the only force acting on it is the tension in the cord. Since the surface is frictionless, there is no friction force. Therefore, the net force on block 1 is equal to the tension in the cord.

For block 2 (m2 = 1.9 kg), the force of gravity is acting downward, and the tension in the cord is acting upward. The net force on block 2 is equal to the difference between these two forces.

Using Newton's second law (F = ma), we can write the following equations:

For block 1: T = m1 * a

For block 2: m2 * g - T = m2 * a

Where T is the tension in the cord and g is the acceleration due to gravity.

Now, we can solve these equations simultaneously to find the acceleration (a).

Let's substitute the first equation into the second equation to eliminate T:

m2 * g - (m1 * a) = m2 * a

Simplifying this equation:

m2 * g = (m1 + m2) * a

a = (m2 * g) / (m1 + m2)

Now, we can substitute the given values:

m1 = 3.2 kg
m2 = 1.9 kg
g = 9.8 m/s^2

a = (1.9 kg * 9.8 m/s^2) / (3.2 kg + 1.9 kg)
a ≈ 4.42 m/s^2

So, the acceleration of the two blocks after they are released is approximately 4.42 m/s^2.

Now, let's move on to the rest of the questions.

(b) To find the velocity of the first block 1.8 s after the release, we can use the equation of motion:

v = u + a * t

Where v is the final velocity, u is the initial velocity (which is 0 in this case since the blocks were released from rest), a is the acceleration, and t is the time.

Substituting the given values:

u = 0 m/s
a = 4.42 m/s^2
t = 1.8 s

v = 0 + (4.42 m/s^2) * (1.8 s)
v ≈ 7.94 m/s

So, the velocity of the first block 1.8 s after the release is approximately 7.94 m/s to the right.

(c) To find the distance block 1 has moved during the 1.8 s interval, we can use the equation:

s = u * t + (1/2) * a * t^2

Where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Substituting the given values:

u = 0 m/s
a = 4.42 m/s^2
t = 1.8 s

s = 0 * (1.8 s) + (1/2) * (4.42 m/s^2) * (1.8 s)^2
s ≈ 7.34 m

So, block 1 has moved approximately 7.34 m during the 1.8 s interval.

(d) To find the displacement of the blocks from their initial positions 0.35 s after they are released, we can use the equation:

s = u * t + (1/2) * a * t^2

Again, substituting the given values:

u = 0 m/s
a = 4.42 m/s^2
t = 0.35 s

s = 0 * (0.35 s) + (1/2) * (4.42 m/s^2) * (0.35 s)^2
s ≈ 0.273 m

So, the displacement of the blocks from their initial positions 0.35 s after they are released is approximately 0.273 m to the right for both blocks.