Two blocks are connected by a very light string passing over a massless and frictionless pulley (the figure ). Traveling at constant speed, the 20.0- block moves 70.0 to the right and the 12.0- block moves 70.0 downward.During this process, how much work is done on the 12.0- block by gravity? During this process, how much work is done on the 12.0- block by the tension in the string? During this process, how much work is done on the 20.0- block by gravity? During this process, how much work is done on the 20.0- block by the tension in the string?
To determine the work done on each block by gravity and tension, we first need to find the displacement and the angles involved.
Let's consider the diagram shown in the figure.
```
m1 = 20.0 kg m2 = 12.0 kg
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│ │
│ │
│ │
```
Given:
- Displacement of m1, Δx1 = 70.0 m to the right
- Displacement of m2, Δx2 = 70.0 m downward
First, let's determine the work done on the 12.0-kg block (m2) by gravity.
The work done by gravity is given by the formula:
Work_gravity = force_gravity * displacement * cosθ
Here, force_gravity = mass * gravity, where gravity = 9.8 m/s^2.
Substituting the values:
- mass (m2) = 12.0 kg
- displacement (Δx2) = 70.0 m
- θ = 90° (since the motion is downward)
Work_gravity = (12.0 kg * 9.8 m/s^2) * (70.0 m) * cos(90°)
Work_gravity = 0 J (since cos(90°) = 0)
Therefore, the work done on the 12.0-kg block by gravity is 0 J.
Next, let's calculate the work done on the 12.0-kg block by the tension in the string.
The tension in the string (T) applies a force in the upward direction, but since the displacement (Δx2) is in the downward direction, the angle between the force and displacement is 180°.
The work done by the tension is given by the formula:
Work_tension = force_tension * displacement * cosθ
Here, force_tension = Tension
Substituting the values:
- Tension (T) = T (same for both blocks)
- displacement (Δx2) = 70.0 m
- θ = 180°
Work_tension = T * (70.0 m) * cos(180°)
Work_tension = -70.0T J
Therefore, the work done on the 12.0-kg block by the tension in the string is -70.0T J.
Moving on to the work done on the 20.0-kg block (m1) by gravity.
The work done by gravity is given by the same formula:
Work_gravity = force_gravity * displacement * cosθ
Substituting the values:
- mass (m1) = 20.0 kg
- displacement (Δx1) = 70.0 m
- θ = 180°
Work_gravity = (20.0 kg * 9.8 m/s^2) * (70.0 m) * cos(180°)
Work_gravity = -13720 J
Therefore, the work done on the 20.0-kg block by gravity is -13720 J.
Finally, let's calculate the work done on the 20.0-kg block by the tension in the string.
Using the same formula:
Work_tension = force_tension * displacement * cosθ
Substituting the values:
- Tension (T) = T (same for both blocks)
- displacement (Δx1) = 70.0 m
- θ = 0° (since the motion is horizontal)
Work_tension = T * (70.0 m) * cos(0°)
Work_tension = 70.0T J
Therefore, the work done on the 20.0-kg block by the tension in the string is 70.0T J.
To find the work done on each block, we need to use the formula:
Work = Force * Distance * cos(theta)
where Force is the magnitude of the force exerted on the object, Distance is the displacement of the object, and theta is the angle between the force and the displacement vectors.
For the 12.0 kg block:
1. Work done by gravity:
The force of gravity acting on an object is given by the equation F = m * g, where m is the mass of the object and g is the acceleration due to gravity. In this case, since the block moves vertically downward, the displacement is also in the same direction as the force of gravity.
To calculate the work done by gravity on the 12.0 kg block, we use the formula:
Work = Force * Distance
Where the force is the weight of the block (m * g) and the distance is 70.0 units downward.
2. Work done by the tension in the string:
The tension in the string acts at an angle of 90 degrees to the displacement of the block. Therefore, the angle between the tension force and the displacement is 90 degrees. Since cos(90 degrees) = 0, the work done by the tension is zero.
For the 20.0 kg block:
1. Work done by gravity:
The force of gravity acting on an object is given by the equation F = m * g, where m is the mass of the object and g is the acceleration due to gravity. In this case, the block moves horizontally to the right, so the force of gravity is perpendicular to the displacement.
Since the angle between the force of gravity and the displacement is 90 degrees, cos(90 degrees) = 0. Therefore, the work done by gravity on the 20.0 kg block is zero.
2. Work done by the tension in the string:
The tension in the string acts at an angle of 180 degrees to the displacement of the block. Therefore, the angle between the tension force and the displacement is 180 degrees. Since cos(180 degrees) = -1, the work done by the tension is negative.
To calculate the work done by the tension on the 20.0 kg block, we use the formula:
Work = Force * Distance * cos(180 degrees)
The force is the tension in the string, and the distance is 70.0 units to the right.
So, in summary:
- Work done on the 12.0 kg block by gravity: m * g * Distance
- Work done on the 12.0 kg block by tension: 0
- Work done on the 20.0 kg block by gravity: 0
- Work done on the 20.0 kg block by tension: -Tension * Distance