A block of mass m1 = 4.1 kg rests on a frictionless horizontal surface. A second block of mass m2 = 1.7 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: What is magnitude and direction?

block 2: What is magnitude and direction?

(b) What is the velocity of the first block 1.4 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?
What is magnitude and direction?

(c) How far has block 1 moved during the 1.4-s interval? in meters

(d) What is the displacement of the blocks from their initial positions 0.50 s after they are released? block 1: Find magnitude and direction

block 2: find magnitude and direction

To solve this problem, we can use Newton's second law of motion and the concept of forces and accelerations.

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

To find the acceleration, we first need to determine the net force acting on each block.

For block 1:
Since it is on a frictionless surface and no external forces are acting on it, its net force is the tension in the string pulling it upward.
Tension = m2 * g (where g is the acceleration due to gravity)
Tension = 1.7 kg * 9.8 m/s^2 = 16.66 N (upward)
The net force on block 1 is equal to the tension.
Net force on block 1 = 16.66 N (upward)

Using Newton's second law, F = m * a (where F is the net force, m is the mass, and a is the acceleration), we can find the acceleration of block 1:
16.66 N = 4.1 kg * a1
a1 = 16.66 N / 4.1 kg = 4.06 m/s^2 (upward)

Therefore, the acceleration of block 1 is 4.06 m/s^2, directed upwards.

For block 2:
The net force on block 2 is the force due to gravity acting downward minus the tension in the string pulling it upward.
Force due to gravity = m2 * g = 1.7 kg * 9.8 m/s^2 = 16.66 N (downward)
Net force on block 2 = Force due to gravity - Tension
Net force on block 2 = 16.66 N - 16.66 N = 0 N (no net force)

Since block 2 has no net force acting on it, its acceleration is zero.
Therefore, the acceleration of block 2 is zero.

(b) What is the velocity of the first block 1.4 s after the release of the blocks?

To find the velocity of block 1, we can use the equation of motion:
v = u + a * t (where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time)

In this case, the initial velocity is zero because the block starts from rest, and the acceleration is 4.06 m/s^2 upwards (as found in part a).

v1 = 0 + 4.06 m/s^2 * 1.4 s = 5.68 m/s (upward)

Therefore, the velocity of block 1 1.4 s after the release of the blocks is 5.68 m/s, directed upwards.

(c) How far has block 1 moved during the 1.4 s interval?

To find the distance traveled by block 1, we can use the equation of motion:
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)

In this case, the initial velocity is zero, and the acceleration is 4.06 m/s^2 upwards (as found in part a).

s1 = 0 + (1/2) * 4.06 m/s^2 * (1.4 s)^2 = 4.6 m

Therefore, block 1 has moved a distance of 4.6 meters during the 1.4 s interval.

(d) What is the displacement of the blocks from their initial positions 0.50 s after they are released?

To find the displacement, we need to consider the motion of both blocks.

For block 1:
The initial velocity is zero, and the acceleration is 4.06 m/s^2 upwards (as found in part a).

s1 = 0 + (1/2) * 4.06 m/s^2 * (0.50 s)^2 = 0.255 m (upward)

Therefore, the displacement of block 1 from its initial position 0.50 s after release is 0.255 meters, directed upward.

For block 2:
Since the acceleration of block 2 is zero (as found in part a), it does not move vertically.

Therefore, the displacement of block 2 from its initial position 0.50 s after release is zero.

To solve this problem, we need to apply Newton's laws of motion. Let's break it down step by step.

(a) acceleration of the two blocks after they are released:

To find the acceleration, we'll begin by considering the forces acting on each block individually.

For block 1:
- The only force acting on it is the tension in the cord.
- The tension in the cord is equal to the weight of block 2, m2 * g, where g is the acceleration due to gravity.
- Since there is no friction, the net force on block 1 is equal to the tension.
- Applying Newton's second law, F = ma, we have m1 * a = m2 * g.
- Rearranging the equation, we find the acceleration: a = (m2 * g) / m1.

For block 2:
- The only force acting on it is its weight, m2 * g.
- There is an upward force due to tension, which is equal to the weight of block 1.
- So, the net force on block 2 is m2 * g - m1 * g.
- Applying Newton's second law, we have m2 * a = m2 * g - m1 * g.
- Rearranging the equation, we find the acceleration: a = (m2 - m1) * g / m2.

Substituting the given values, m1 = 4.1 kg, m2 = 1.7 kg, and g = 9.8 m/s^2, we can calculate the accelerations.

For block 1: a = (1.7 kg * 9.8 m/s^2) / 4.1 kg = 4.06 m/s^2 (towards the right)

For block 2: a = (1.7 kg - 4.1 kg) * 9.8 m/s^2 / 1.7 kg = -21.18 m/s^2 (downwards)

(b) velocity of the first block 1.4 s after the release:

To find the velocity of block 1, we can use the formula v = u + at, where u is the initial velocity, a is the acceleration, and t is the time.

We know that block 1 is initially at rest, so u = 0 m/s. The acceleration is given as 4.06 m/s^2. Plugging these values in, we get:

v = 0 + (4.06 m/s^2) * (1.4 s) = 5.68 m/s (to the right)

(c) distance moved by block 1 during the 1.4 s interval:

To find the distance moved by block 1, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, and t is the time.

We already found that the initial velocity u = 0 m/s, and the acceleration a = 4.06 m/s^2. Plugging these values in, we get:

s = 0 + (1/2) * (4.06 m/s^2) * (1.4 s)^2 = 4.47 m

Therefore, block 1 moved a distance of 4.47 meters during the 1.4 s interval.

(d) displacement of the blocks from their initial positions 0.50 s after release:

To find the displacement, we'll calculate the distance traveled by each block and consider their respective directions.

For block 1:
- We can use the equation s = ut + (1/2)at^2, similar to the previous calculation.
- The initial velocity u = 0 m/s, and the acceleration a = 4.06 m/s^2.
- Plugging in the values, we have s = 0 + (1/2) * (4.06 m/s^2) * (0.5 s)^2 = 0.51 m to the right.

For block 2:
- We'll use the equation s = ut + (1/2)at^2, with u = 0 m/s (since the block starts from rest) and a = -21.18 m/s^2 (downwards).
- Plugging in the values, we have s = 0 + (1/2) * (-21.18 m/s^2) * (0.5 s)^2 = -0.66 m downwards.

Therefore, the displacement of block 1 is 0.51 meters to the right, and the displacement of block 2 is 0.66 meters downwards.

a). The direction should be obvious. Block m1 slides horizontally and Block m2 falls vertically. For the acceleration, a, write free body Newton's-law equations for each blcok separately, with the connecting cord tension T acting on both. Eliminate T and solve for a. Both blocks accelerate at the same rate, but in different directions.

b) velocity = a t (horizontal direction)

c) Displacement = (1/2) a t^2

d) Use the same formula as (c) but different time t

Do the calculations yourself and learn the method.