1. Oct 27, 2004

### Nikki196

I have a few questions I've been having troubles with, and I was hoping someone could help me out...
1) A machine does 4000 J of work in accelerating a 20 kg object, intially at rest, at 3.0 m/s2 for 5.0s determine
a) The efficiency of the machine
b) The power output of the machine

2)A block and tackle is used to raise a 100kg load. If the maximum applied force is 220N, how many strands of rope must support the moveable block?

3)A wrench is used to turn a bolt of 6.0mm diameter. The centre of the bolt head is 0.2m from the applied force.
a) what is the IMA of this machine?
b) If the efficiency is 50%, what is the AMA of the machine?

4) A ramp is 12m in length and has a grade of 25%
a)If friction is neglected, what force is required to push a 120-kg box up the ramp?
b) If the actual force required is 490 N, determine the efficiency of the ramp.

If someone could help me it would be awesome!
Nikki196

2. Oct 27, 2004

### Diane_

That's an awful lot of questions. Let me concentrate on the first.

You have a machine doing a certain amount of work on an object. One thing you want to ask yourself is "where does the energy go" - remembering, of course, that work is energy transferred by mechanical means.

One answer to the question is, obviously, into the speed the object acquires. The energy of motion is called kinetic energy - some of the work done by the machine shows up as kinetic energy. How much? Well, you have a formula to determine kinetic energy. In that formula, you'll have to know the speed of the object, but that's easy to determine - if an object accelerates from rest at 3.0 m/s^2 for 5.0 s, how fast does it end up travelling? Once you have that, you'll have the information necessary to find the kinetic energy.

You'll find that it's less than 4000 J. OK - so what happened to the rest? Well, given the problem, the only thing that could have happened is inefficiency - overcoming friction (hence generating heat), making noise (also eventually generating heat), things like that. How much was lost to that? Well, what's the difference between 4000 J and the kinetic energy of the object?

Once you know that, you can determine the efficiency of the machine. Remember, it would be perfectly efficient if all 4000 J went into "useful" work - in this case, "useful" being defined as kinetic energy. So, take the "useful" work - whatever the kinetic energy was - and divide by the work input, in this case 4000 J. Multiply the result by 100 to turn it into a percent, and Bob's your uncle.

The power output is easy. It did 4000 J of work in 5.0 s. Power is work/time. Just division. And remember - watt is the unit of power. (This is a lot funnier if you hear it instead of read it.)