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ICSE Class 10 Answered

What is the power of a machinethat can lift a load of 750N through 15m in 5 sec? What must have been the effort applied? Obtain the M.A. and V.R. of this machine. Assume the machine to be an ideal one and take the distance moved by the effort as 45m.
Asked by bdroy_2014 | 25 Jan, 2018, 01:16: PM
answered-by-expert Expert Answer
Power of machine to lift a load W to height h in t seconds = W×h×t
(units of W and h are Newton and metre respectively)
 
Hence power = 750×15×5 = 56.25 kW
 
for ideal machine,
 
effort × distance through which effort acts = load × distance to move the load
 
effort × 45 = 750 × 15;  hence effort = 250 N
 
Mechanical advantage = load/effort = 750/250 = 3
 
velocity ratio = distance moved by effort/ distance moved for load = 45/15 = 3
Answered by Thiyagarajan K | 25 Jan, 2018, 01:54: PM

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