A black and tackle with a velocity ratio of 5 is used to raise a mass of 25kg through a vertical distance of 40cm at a steady rate.if the effort is equal to 60N, determine
a. The distance moved by the effort
b. The work done by the effort in lifting the load
40 * 5
60 * 5 * 0.40 Joules
I want to know more about this calculation
To solve this problem, we need to first understand some key concepts related to a block and tackle system and work.
A block and tackle is a type of mechanical system that consists of a rope or chain, a set of pulleys or sheaves, and a load. The pulleys are arranged in such a way that they are attached to both the load and the effort applied to lift the load. The mechanical advantage, or velocity ratio, of a block and tackle system is the ratio of the load lifted to the effort applied.
In this case, the velocity ratio is given as 5, meaning that for every 5 units of effort applied, the load is lifted by 1 unit. The load, in this case, is a mass of 25 kg.
Let's find the distance moved by the effort (a) and the work done by the effort in lifting the load (b):
a. Distance moved by the effort:
The distance moved by the effort is equal to the product of the velocity ratio and the distance moved by the load. In this case, the distance moved by the load is given as 40 cm.
Distance moved by the effort = Velocity ratio * Distance moved by the load
= 5 * 40 cm
Therefore, the distance moved by the effort is 200 cm.
b. Work done by the effort:
Work is defined as the product of force and distance. In this case, the force applied by the effort is given as 60 N.
Work done by the effort = Force * Distance moved by the effort
= 60 N * 200 cm
However, we need to convert the distance from centimeters to meters, as the SI unit for work is joules (J).
1 meter = 100 centimeters
200 cm * (1 meter / 100 cm) = 2 meters
Work done by the effort = 60 N * 2 meters
= 120 J
Therefore, the work done by the effort in lifting the load is 120 joules.