011 (part 1 of 5) 10.0 points
Two objects are connected by a string of negligible mass. The 12 kg block is placed on a
smooth table top 2.4 m above the floor, and
the 9 kg block hangs over the edge of the table. The 9 kg block is then released from rest
at a distance of 1.2 m above the floor at time
t = 0.
12 kg
9 kg
2.4 m
1.2 m
Determine the acceleration of the 9 kg block
as it descends. The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m/s
2
How long dies it take the 9 kg block to strike
the floor?
Answer in units of s
013 (part 3 of 5) 10.0 points
Describe the motion of the 12 kg block from
the time t = 0 to the time when the 9 kg block
strikes the floor.
1. It moves with a decreasing acceleration.
2. It moves with an increasing acceleration.
3. It moves with a constant acceleration.
4. It moves with a constant speed.
5. It stays still.
014 (part 4 of 5) 10.0 points
Describe the motion of the 12 kg block from
the time the 9 kg block strikes the floor to the
time the 12 kg block leaves the table.
1. It stays still.
2. It moves with a constant speed.
3. It moves with a decreasing acceleration.
4. It moves with a constant acceleration.
5. It moves with an increasing acceleration.
015 (part 5 of 5) 10.0 points
Determine the distance between the landing
points of the two blocks.
Answer in units of m
First, let's find the acceleration of the 9 kg block as it descends.
The downward force acting on the 9 kg block is its weight, which can be calculated as:
F = m * g
F = 9 kg * 9.8 m/s²
F = 88.2 N
Since the string connecting the two blocks is massless, the tension in the string is equal to the downward force acting on the 9 kg block. Therefore, the tension in the string is also 88.2 N. This tension force acts horizontally on the 12 kg block, causing it to accelerate.
a = F / m
a = 88.2 N / 12 kg
a ≈ 7.35 m/s²
Now, let's find out how long it takes for the 9 kg block to strike the floor.
We can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time²
where distance is 1.2 m (the distance the 9 kg block falls), initial velocity is 0 (since the block starts from rest), acceleration is 9.8 m/s² (acceleration due to gravity), and time is what we're solving for.
1.2 m = 0 + 0.5 * 9.8 m/s² * time²
Rearrange and solve for time:
time² = 1.2 m / (0.5 * 9.8 m/s²)
time² ≈ 0.2449 s²
time ≈ 0.495 s
Now let's describe the motion of the 12 kg block between t = 0 and when the 9 kg block strikes the floor. As there is a constant force of tension acting on it and no other forces in the horizontal direction, it moves with a constant acceleration (answer 3).
When the 9 kg block strikes the floor, the tension in the string goes to zero, so there are no longer any horizontal forces acting on the 12 kg block. Therefore, it will continue to move with a constant speed (answer 2).
Finally, let's find the distance between the landing points of the two blocks. During the time it takes for the 9 kg block to fall (0.495 s), the 12 kg block is moving with a constant acceleration of 7.35 m/s². We can use the equation for distance traveled under constant acceleration:
distance = initial velocity * time + 0.5 * acceleration * time²
The 12 kg block starts from rest, so its initial velocity is 0. Plug in the values:
distance = 0 + 0.5 * 7.35 m/s² * (0.495 s)²
distance ≈ 0.902 m
The distance between the landing points of the two blocks is approximately 0.902 m.
To find the acceleration of the 9 kg block as it descends, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Since the 9 kg block is hanging freely and there is no other force acting on it except for gravity, the net force on the block is simply its weight. The weight of an object can be calculated as the mass of the object multiplied by the acceleration due to gravity.
Weight = mass * acceleration due to gravity
Weight = 9 kg * 9.8 m/s^2
Weight = 88.2 N
Now, using the net force formula, we have:
Net force = mass * acceleration
88.2 N = 9 kg * acceleration
Solving for acceleration:
acceleration = 88.2 N / 9 kg
acceleration ≈ 9.8 m/s^2
So, the acceleration of the 9 kg block as it descends is approximately 9.8 m/s^2.
To find how long it takes the 9 kg block to strike the floor, we can use the second equation of motion:
Δy = v0 * t + (1/2) * a * t^2
Where:
Δy = displacement (1.2 m)
v0 = initial velocity (0 m/s, since the block is released from rest)
a = acceleration (9.8 m/s^2, as found above)
t = time
Rearranging the equation:
(1/2) * a * t^2 + v0 * t - Δy = 0
Solving this quadratic equation for t, we can use the quadratic formula:
t = (-v0 ± sqrt(v0^2 - 4 * (1/2) * a * -Δy)) / (2 * (1/2) * a)
Plugging in the values:
t = (-0 ± sqrt(0^2 - 4 * (1/2) * 9.8 * -1.2)) / (2 * (1/2) * 9.8)
Simplifying:
t = sqrt(2.4 / 4.9)
t ≈ 0.61 seconds
So, it takes approximately 0.61 seconds for the 9 kg block to strike the floor.
Describing the motion of the 12 kg block from the time t = 0 to the time when the 9 kg block strikes the floor, the correct answer is:
3. It moves with a constant acceleration.
Describing the motion of the 12 kg block from the time the 9 kg block strikes the floor to the time the 12 kg block leaves the table, the correct answer is:
5. It moves with an increasing acceleration.
To determine the distance between the landing points of the two blocks, we need to find the horizontal distance traveled by the 12 kg block while the 9 kg block fell.
The time it takes for the 9 kg block to fall to the floor is approximately 0.61 seconds, as found earlier.
In this time, the 12 kg block is subject to a constant horizontal acceleration and is moving with an increasing speed.
Using the third equation of motion:
Δx = v0 * t + (1/2) * a * t^2
Where:
Δx = horizontal distance
v0 = initial velocity (0 m/s, since the block is at rest initially)
a = horizontal acceleration
t = time (0.61 seconds)
Since the 12 kg block moves horizontally with a constant acceleration, we can use the formula:
Δx = (1/2) * a * t^2
Solving for Δx:
Δx = (1/2) * a * t^2
Δx = (1/2) * (0.61 s)^2 * a
The acceleration can be found using the formula:
a = g * sin(θ)
Where:
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of the incline, which is equal to the arctan(2.4 m/1.2 m)
θ = arctan(2.4 m/1.2 m)
θ ≈ 63.43 degrees
a = g * sin(63.43 degrees)
a ≈ 9.8 m/s^2 * 0.891
a ≈ 8.72 m/s^2
Plugging in the values:
Δx = (1/2) * (0.61 s)^2 * 8.72 m/s^2
Δx ≈ 0.126 m
So, the distance between the landing points of the two blocks is approximately 0.126 meters.
To find the acceleration of the 9 kg block as it descends, we can use the concept of gravity and the formula for acceleration.
Since the 9 kg block is released from rest, its initial velocity is 0 m/s. The only force acting on it is gravity, so the acceleration of the block will be equal to the acceleration due to gravity, which is 9.8 m/s^2. Therefore, the acceleration of the 9 kg block as it descends is 9.8 m/s^2.
To find the time it takes for the 9 kg block to strike the floor, we can use the formula for displacement.
The displacement of the 9 kg block is 1.2 m (the distance it is released above the floor), and we need to find the time it takes to travel this distance. Since the block is only under the influence of gravity, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity (0 m/s)
a = acceleration (9.8 m/s^2)
t = time
Plugging in the values, we get:
1.2 = 0 + (1/2)(9.8)t^2
Simplifying further:
1.2 = 4.9t^2
Dividing by 4.9:
t^2 = 1.2/4.9
Taking the square root of both sides:
t = √(1.2/4.9)
Using a calculator, we can find that t ≈ 0.3809 seconds.
Therefore, it takes approximately 0.3809 seconds for the 9 kg block to strike the floor.
Now let's move on to describing the motion of the 12 kg block:
From the time t = 0 to the time when the 9 kg block strikes the floor, the 12 kg block moves with a constant acceleration. This is because the only force acting on the 12 kg block is the tension in the string, which is proportional to the acceleration of the 9 kg block.
So, the answer to question 013 is: 3. It moves with a constant acceleration.
Next, let's describe the motion of the 12 kg block from the time the 9 kg block strikes the floor to the time the 12 kg block leaves the table:
After the 9 kg block strikes the floor, there is no longer a tension force in the string, as it has been cut or released. Therefore, the 12 kg block will no longer have any horizontal force acting on it. It will continue moving with its constant horizontal velocity until it leaves the table.
So, the answer to question 014 is: 2. It moves with a constant speed.
Finally, let's find the distance between the landing points of the two blocks:
We know that the 9 kg block falls a distance of 1.2 m. To find the distance between the landing points, we need to find the horizontal displacement of the 12 kg block.
The time it takes for the 9 kg block to fall is the same as the time it takes for the 12 kg block to move horizontally. So, using the time we calculated earlier (t ≈ 0.3809 s), we can find the horizontal displacement of the 12 kg block:
d = vt
Since the horizontal velocity of the 12 kg block is constant, we can use the initial horizontal velocity given by the string before it was cut or released. Since the blocks were initially at rest and the string is of negligible mass, both blocks move as one until the string is cut. Therefore, the initial horizontal velocity of the 12 kg block is 0 m/s.
Plugging in the values, we get:
d = 0 * 0.3809
d = 0
So, the distance between the landing points of the two blocks is 0 meters.
Therefore, the answer to question 015 is: 0 m.