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For a machine with an efficiency of 80%, what is the work output of the machine if its work input is 2,000 joules?

would i divide 2000/.80? which gives me 2500?

If a machine requires an effort force of 750 newtons applied over a distance of 2 meters, over what resistance distance is the machine acting if the resistance force has a value of 150 newtons? Assume that it is an ideal machine. (HINT: Use the equation Fe x de = F r x dr)
A 5 meters
B 10 meters
C 150 meters
D 300 meters

can you help me set up the equation cause im lossttt! please(:

  • science -

    No to your first question. You can't get more work out than you put in.

    Second question: For a 100% efficient machine, the product of force x distance is the same for the input (effort) as the output (resistance). That is the equation they told you to use. So, use it. You want to solve for the resistance distance dr

    dr = Fe*de/Fr = 750*2/150 = ? meters

  • DRWLS -

    so would i mult. 2000* .80 which gives me 1600?

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