Questions: A rope is threaded such that it is supporting a movable pulley directly in the middle between a fixed anchor point 2.8 m high on the wall and a fixed pulley at the same height a distance 4.0 m away. After going over the fixed pulley, the rope angles downward, reaching the ground a distance of 7.0 m away from the pulley. A mass of 36 kg is hanging from the movable pulley, and is initially 0.8 m off the ground. If the end of the rope that reaches the ground is pulled 0.3 m further away from the pulleys, what will the change in the mass' gravitational potential energy be? Answer to one decimal place.

Solution Steps

We need to determine the change in gravitational potential energy of a mass hanging from a movable pulley when the end of the rope is pulled 0.3 m further away from the pulleys.

• Height of the fixed anchor point and fixed pulley: \(2.8 \, \text{m}\)
• Horizontal distance between the fixed anchor point and the fixed pulley: \(4.0 \, \text{m}\)
• Horizontal distance from the fixed pulley to the ground point where the rope reaches: \(7.0 \, \text{m}\)
• Mass hanging from the movable pulley: \(36 \, \text{kg}\)
• Initial height of the mass above the ground: \(0.8 \, \text{m}\)
• Distance the rope is pulled further away: \(0.3 \, \text{m}\)

• The initial length of the rope from the fixed anchor point to the movable pulley: \[ L_1 = \sqrt{(2.8 \, \text{m})^2 + (2.0 \, \text{m})^2} = \sqrt{7.84 + 4.0} = \sqrt{11.84} \approx 3.44 \, \text{m} \]
• The initial length of the rope from the fixed pulley to the ground point: \[ L_2 = \sqrt{(2.8 \, \text{m})^2 + (7.0 \, \text{m})^2} = \sqrt{7.84 + 49.0} = \sqrt{56.84} \approx 7.54 \, \text{m} \]

When the rope is pulled 0.3 m further away:

• The new horizontal distance from the fixed pulley to the ground point: \[ 7.0 \, \text{m} + 0.3 \, \text{m} = 7.3 \, \text{m} \]
• The new length of the rope from the fixed pulley to the ground point: \[ L_2' = \sqrt{(2.8 \, \text{m})^2 + (7.3 \, \text{m})^2} = \sqrt{7.84 + 53.29} = \sqrt{61.13} \approx 7.82 \, \text{m} \]

• Change in length of the rope segment from the fixed pulley to the ground point: \[ \Delta L_2 = L_2' - L_2 = 7.82 \, \text{m} - 7.54 \, \text{m} = 0.28 \, \text{m} \]
• Since the rope is pulled 0.3 m, the movable pulley will move up by half of this change: \[ \Delta h = \frac{0.28 \, \text{m}}{2} = 0.14 \, \text{m} \]

• Change in height of the mass: \[ \Delta h = 0.14 \, \text{m} \]
• Change in gravitational potential energy: \[ \Delta U = m \cdot g \cdot \Delta h = 36 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.14 \, \text{m} \approx 49.5 \, \text{J} \]

Final Answer