Question

A system of pulley has a velocity ratio of 6 and mechanical advantage of 5 the system is used to raise a load of 3000N through 10m calculate the workdone on load

Answer 1

Work done on the load = force applied x distance moved in the direction of force

In this case, the force applied is the effort force, which is the force required to lift the load, and the distance moved is the height through which the load is lifted.

The mechanical advantage of the pulley system is given by:

MA = Load / Effort

Since the MA is 5, the effort force required to lift the load is:

Effort = Load / MA

Effort = 3000N / 5

Effort = 600N

The velocity ratio of the pulley system is given by:

Velocity ratio = distance moved by effort / distance moved by load

Since the velocity ratio is 6, the distance moved by the effort force is:

Distance moved by effort = 6 x Distance moved by load

Distance moved by load = 10m (given)

Distance moved by effort = 6 x 10m

Distance moved by effort = 60m

Therefore, the work done on the load is:

Work done on load = Effort x Distance moved by load

Work done on load = 600N x 10m

Work done on load = 6000 Joules

Therefore, the work done on the load is 6000 Joules.