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.
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
1 answer
1 answer