A load of 500N is raised 0.2m by a machine in which an effort of 150N moves 1m.What is a) the work done on the load b)the work done by the effort c)the efficiency
As you should have learned by now, work is force x distance. Use that relationship to answer (a) and (b).
For the efficiency (c), divide load work (from a) by effort work (from b). In this case (c), the answer is 2/3.
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To find the work done on the load, we can use the formula:
Work = Force × Distance
a) The work done on the load can be calculated as:
Work = Force × Distance
= 500N × 0.2m
= 100 Joules
Therefore, the work done on the load is 100 Joules.
b) The work done by the effort can also be calculated using the same formula:
Work = Force × Distance
= 150N × 1m
= 150 Joules
Therefore, the work done by the effort is 150 Joules.
c) Efficiency can be calculated using the formula:
Efficiency = (useful output energy / input energy) × 100%
In this case, the input energy is the work done by the effort (150 Joules) and the useful output energy is the work done on the load (100 Joules).
Efficiency = (100 Joules / 150 Joules) × 100%
= 66.7%
Therefore, the efficiency is approximately 66.7%.