The two masses on either side of the pulley, 2.2 kg and 3.6 kg, are each initially 1.8 m above the ground, and the massless frictionless pulley is 4.8m above the ground.What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its "launch" speed. Assume the mass doesn't hit the pulley. Ignore the mass of the cord.]
1.8
To determine the maximum height reached by the lighter object, we need to follow these steps:
Step 1: Calculate the acceleration of the system.
Since the pulley is massless and frictionless, the tension in the string will be the same on both sides. Let's denote the acceleration as "a". Considering the forces involved, we can set up the following equation:
T - m₁g = m₁a ---- (Equation 1)
T - m₂g = -m₂a ---- (Equation 2)
Where:
T = tension in the string
m₁ = mass of the lighter object = 2.2 kg
m₂ = mass of the heavier object = 3.6 kg
g = acceleration due to gravity ≈ 9.8 m/s²
Step 2: Solve the system of equations.
Substituting the given values in Equations 1 and 2:
T - 2.2(9.8) = 2.2a ---- (Equation 3)
T - 3.6(9.8) = -3.6a ---- (Equation 4)
Step 3: Find the expression for acceleration.
To solve the system of equations, combine Equation 3 and Equation 4:
T - 2.2(9.8) = 2.2a
T - 3.6(9.8) = -3.6a
Simplifying,
T = 21.56 + 2.2a ----(Equation 5)
T = 35.28 - 3.6a ----(Equation 6)
Since T is the same in both equations,
21.56 + 2.2a = 35.28 - 3.6a
Combine like terms,
5.8a = 13.72
Simplify,
a = 2.37 m/s²
Step 4: Calculate the velocity of the lighter mass when the heavier mass hits the ground.
Using the derived acceleration value, we can determine the velocity of the lighter mass right before the heavier mass hits the ground.
To do this, we can use the kinematic equation:
v² = u² + 2as
Where:
v = final velocity (when the heavier mass hits the ground)
u = initial velocity (at rest, since it starts from rest)
a = acceleration (2.37 m/s²)
s = displacement (distance covered by the lighter mass)
Since the lighter mass travels 1.8 m downwards, the displacement will be negative (-1.8 m).
So, the equation becomes:
v² = 0 + (2 * 2.37 * -1.8)
v² = -8.5724
Taking the square root,
v ≈ -2.93 m/s
(Note: The negative sign represents the downward direction.)
Step 5: Calculate the maximum height reached by the lighter object.
The maximum height can be calculated using the following equation:
v² = u² + 2as
Where:
v = final velocity (0 m/s, when the lighter mass reaches the maximum height)
u = initial velocity (-2.93 m/s, determined earlier)
a = acceleration (2.37 m/s²)
s = displacement (unknown, the maximum height reached)
The equation becomes:
0 = (-2.93)² + 2 * 2.37 * s
0 = 8.5849 + 4.74s
Rearranging the equation:
4.74s = -8.5849
Dividing by 4.74:
s ≈ -1.81 m
(Note: The negative sign represents the downward direction.)
Step 6: Determine the maximum height above the ground.
Since the initial height was 1.8 m above the ground, and the displacement measured in Step 5 was -1.81 m (downward), we can subtract the displacement from the initial height to find the maximum height:
Maximum Height = Initial Height + Displacement
Maximum Height = 1.8 m - (-1.81 m)
Maximum Height ≈ 3.61 m
Therefore, the lighter object reaches a maximum height of approximately 3.61 meters above the ground after the system is released.
To find the maximum height reached by the lighter object after the system is released, we need to consider the conservation of mechanical energy.
Here are the steps to solve the problem:
1. Find the acceleration of the system:
- The two masses are connected by a massless and frictionless pulley, so their accelerations are equal in magnitude.
- Let's assume the acceleration of the system to be 'a'.
- Using Newton's second law, the net force on each mass can be related to their respective masses:
- For the heavier mass (3.6 kg): F₁ = m₁ * g - m₁ * a
- For the lighter mass (2.2 kg): F₂ = m₂ * g + m₂ * a
- Since the two masses are connected by a massless string, their magnitudes of acceleration are equal: a = a₁ = a₂
- We can solve these equations simultaneously to find the value of 'a'.
2. Determine the velocity of the lighter mass when the heavier one hits the ground:
- Since we're assuming the lighter mass doesn't hit the pulley, it will reach its maximum height at the moment the heavier mass hits the ground.
- At this point, the lighter mass will act as a projectile, launched with an initial velocity.
- Since we know the acceleration 'a' (from step 1), we can find the time taken for the heavier mass to hit the ground: t = √(2 * h / a), where 'h' is the height from which the lighter mass was released (1.8 m in this case).
- With the value of 't', we can use kinematic equations (assuming initial vertical velocity is zero) to find the velocity of the lighter mass at this moment: v = a * t.
3. Determine the maximum height reached by the lighter mass:
- Once we have the 'launch' speed of the lighter mass, we can use the conservation of mechanical energy.
- The lighter mass loses its initial kinetic energy but gains potential energy as it rises.
- At the maximum height, all of its initial kinetic energy will be converted to potential energy (assuming no energy losses due to friction or other factors).
- We can write the equation: m₂ * (v² / 2) = m₂ * g * h_max
- Solving for 'h_max', we can find the maximum height reached by the lighter mass.
By following these steps, you can calculate the maximum height reached by the lighter mass after the system is released.