Two blocks are connected by a string and are pulled vertically upward by a force of 165 N applied to the upper block.

(here is a rough diagram)
^ (F=165 N)
I
[block] ( <- 2.00 kg)
I
[block] ( <- 1.00 kg

(a) Find the tension T in the string connecting the blocks.
(b) If the blocks start from rest, what is their velocity after having moved a distance of 10.0 cm?

I have the answers to both from the back of the book, but do not know how to figure them out offhand.
A= 55 N
B= 3.01 m/s^2

I tried to find the acceleration, though I believe its not right. Here's what I did.

(m2= mass 2, m1= mass 1)
a= (m2-m1)g/(m2+m1)
a= (2.00-1.00)(9.80m/s^2)/(2.00+1.00)
a= 9.80/3.00
a= 3.267

that's all i have so far, any help would be appreciated. Thanks for your time!

bozo

To solve this problem, we can use Newton's second law of motion and the concept of tension in the string. Let's go through the steps for each part of the question:

(a) Finding the tension T in the string connecting the blocks:
To determine the tension in the string, we need to consider the forces acting on each block.

For the upper block:
The only force acting on it is the applied force of 165 N. Since the block doesn't move vertically, the net force in the upward direction is zero. Thus, the tension in the string will be equal to the applied force, T = 165 N.

(b) Finding the velocity after moving 10.0 cm:
To find the velocity of the blocks after moving a distance of 10.0 cm, we need to calculate the acceleration first using Newton's second law.

For the system of blocks:
The net force acting on the system is the tension T pulling the blocks upwards. The weight of the blocks (mg) is acting downwards. Using Newton's second law, we have:

ΣF = ma

The net force is T - (m1g + m2g), where g is the acceleration due to gravity. The masses m1 and m2 are given as 1.00 kg and 2.00 kg, respectively.

T - (m1g + m2g) = (m1 + m2)a

Substituting the given values:
165 N - (1.00 kg * 9.80 m/s^2 + 2.00 kg * 9.80 m/s^2) = (1.00 kg + 2.00 kg) * a

165 N - 9.80 N - 19.6 N = 3.00 kg * a

135.2 N = 3.00 kg * a

Now we can solve for the acceleration, a:

a = 135.2 N / 3.00 kg
a ≈ 45.07 m/s^2

Once we have the acceleration, we can use the kinematic equation to find the final velocity, v:

v^2 = u^2 + 2as

Where:
u = initial velocity (0 m/s, as the blocks start from rest)
s = distance moved (10.0 cm = 0.10 m)
a = acceleration (45.07 m/s^2)

Solving for v:

v^2 = 0 + 2 * 45.07 m/s^2 * 0.10 m
v^2 = 9.014 m^2/s^2

v ≈ √(9.014) ≈ 3.00 m/s (rounded to two decimal places)

So, the final velocity of the blocks after moving a distance of 10.0 cm is approximately 3.00 m/s.

Note: There seems to be a discrepancy in the answer for part (b) provided in the book. The correct answer should be approximately 3.00 m/s, rather than 3.01 m/s as mentioned.

To solve this problem, you can start by analyzing the forces acting on each block individually and then using Newton's second law of motion. Let's break it down step by step:

Step 1: Analyzing the forces on the blocks
For each block, there are two forces acting on it: the force of gravity (mg) and the tension force in the string (T).

For the 2.00 kg block:
The force of gravity (mg) is given by:
F_g1 = m1 * g = 2.00 kg * 9.80 m/s^2 = 19.60 N

For the 1.00 kg block:
The force of gravity (mg) is given by:
F_g2 = m2 * g = 1.00 kg * 9.80 m/s^2 = 9.80 N

Step 2: Applying Newton's second law of motion
For each block, Newton's second law states F_net = m * a, where F_net is the net force acting on the block, m is the mass of the block, and a is the acceleration.

For the 2.00 kg block:
The net force is the difference between the tension force (T) pulling it upward and the force of gravity (F_g1) pulling it downward:
F_net1 = T - F_g1 = T - 19.60 N

For the 1.00 kg block:
The net force is the difference between the tension force (T) pulling it upward and the force of gravity (F_g2) pulling it downward:
F_net2 = T - F_g2 = T - 9.80 N

Step 3: Relating the accelerations of both blocks
Since the string pulls both blocks together, their accelerations must be the same. Let's call this common acceleration "a".

Step 4: Solving for acceleration
Using Newton's second law for each block:
F_net1 = m1 * a
T - 19.60 N = 2.00 kg * a [Equation 1]

F_net2 = m2 * a
T - 9.80 N = 1.00 kg * a [Equation 2]

Step 5: Solving for Tension (T)
Simplify and solve for T by combining Equation 1 and Equation 2:
2.00 kg * a + 19.60 N = 1.00 kg * a + 9.80 N

1.00 kg * a = 19.60 N - 9.80 N
1.00 kg * a = 9.80 N

a = 9.80 m/s^2

Now substitute the value of acceleration (a) back into either Equation 1 or Equation 2 to solve for T:
T - 19.60 N = 2.00 kg * 9.80 m/s^2

T - 19.60 N = 19.60 N
T = 19.60 N + 19.60 N
T = 39.20 N

So, the tension in the string connecting the blocks is 39.20 N.

For part (b) of the question, we need to find the final velocity of the blocks after they have moved a distance of 10.0 cm.

Step 1: Convert the distance to meters:
10.0 cm = 0.1 meters

Step 2: Use the equation of motion to find the final velocity (v):
v^2 = u^2 + 2ad

Where:
v = final velocity (to be found)
u = initial velocity (assumed to be zero as the blocks start from rest)
a = acceleration (found earlier to be 9.80 m/s^2)
d = distance (0.1 meters)

v^2 = 0 + 2 * 9.80 m/s^2 * 0.1 m
v^2 = 1.96 m^2/s^2

Taking the square root of both sides:
v = √(1.96 m^2/s^2)
v ≈ 1.40 m/s

Therefore, the final velocity of the blocks after moving a distance of 10.0 cm is approximately 1.40 m/s.