Fig. 33.1 shows a pulley system used to lift a 200 N load. A person pulls with a force F on the rope. The pulley is ideal (no friction, weightless). (a) Sta...
Question 1 Report
Fig. 33.1 shows a pulley system used to lift a 200 N load. A person pulls with a force F on the rope.
The pulley is ideal (no friction, weightless).
(a) State the force F needed to hold the load in equilibrium. [1]
(b) The person pulls the rope down by 2.0 m. State how far the load rises. [1]
(c) Calculate the work done by the person. [2]
(d) Calculate the gravitational potential energy gained by the load. [2]
(e) State why in practice the work done by the person is greater than the GPE gained. [1]
Answer Details
(a) This is a single fixed pulley, which changes the direction of the applied force but provides no mechanical advantage. The force F needed equals the load:
\( F = \textbf{200 N} \) [1]
The pulley simply redirects the force so the person can pull downwards to lift the load upwards.
(b) With a single fixed pulley, the velocity ratio is 1. The rope pulled equals the distance the load rises:
Load rises = 2.0 m [1]
(c) Work done by the person = force applied x distance pulled:
\( W = F \times d = 200 \times 2.0 = \textbf{400 J} \) [2]
(d) The gravitational potential energy gained equals the weight times the height gained. Here the weight is given directly as 200 N:
\( \text{GPE} = Wh = 200 \times 2.0 = \textbf{400 J} \) [2]
For an ideal (frictionless) pulley, the work input equals the GPE gained, confirming 100% efficiency.
(e) In practice, there is friction at the pulley axle, so some of the work done by the person is converted to thermal energy (heat) rather than GPE. The person must therefore do more work than the GPE gained. [1]
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