With the help of a movable block, a 45kg load was evenly lifted to a height of 2m.

With the help of a movable block, a 45kg load was evenly lifted to a height of 2m. To lift the load, a force of 300 N was applied to the end of the cable. Determine the CPL of the movable block.

Task data: m (weight of the load lifted by the movable block) = 45 kg; the lifting of the load was uniform; h (lifting height, distance traveled by the load) = 2 m; F (force applied to the end of the rope) = 300 N.

Reference values: g (acceleration due to gravity) ≈ 10 m / s2.

The efficiency of the used mobile unit is determined by the formula: η = Ap / Az = m * g * h / (F * 2h) = m * g / 2F.

Calculation: η = 45 * 10 / (2 * 300) = 0.75 (75%).

Answer: The efficiency of the used mobile unit is 75%.