How Do You Calculate Physics Machine Efficiency and Power?

If all of that was a bit much, here's a summary:In summary, the first problem involves a machine doing 4000 J of work on an object to accelerate it from rest at 3.0 m/s^2 for 5.0 s. The remaining questions deal with determining the efficiency and power output of the machine, as well as considering where the energy goes and what factors contribute to inefficiency.
  • #1
Nikki196
3
0
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
 
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  • #2
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.)
 
  • #3
,

1) a) To determine the efficiency of the machine, we need to first calculate the input work and output work. Input work is the work done by the machine, which is 4000 J. Output work is the work done on the object, which can be calculated using the formula W = Fd, where W is the work, F is the force and d is the distance. In this case, the force is the weight of the object, which is m x g, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). So, the output work is W = (20 kg)(9.8 m/s^2)(5.0 s) = 980 J. Efficiency is calculated by dividing the output work by the input work and multiplying by 100 to get a percentage. So, the efficiency of the machine is (980 J / 4000 J) x 100 = 24.5%.

b) Power is the rate at which work is done, and it is calculated by dividing work by time. In this case, the power output of the machine is (4000 J / 5.0 s) = 800 watts (or 800 J/s).

2) To determine the number of strands of rope needed, we need to use the formula for mechanical advantage (MA) of a block and tackle system, which is MA = 2 x number of ropes supporting the moveable block. In this case, the MA is 220 N / 100 kg = 2.2. So, 2.2 = 2 x number of ropes, which means that the number of strands of rope needed is 2.2 / 2 = 1.1. Since we can't have a fraction of a rope, we need to round up to the nearest whole number. So, the block and tackle system needs 2 strands of rope to support the moveable block.

3) a) The IMA (ideal mechanical advantage) of a machine is the ratio of the input distance to the output distance. In this case, the input distance is the distance from the applied force to the centre of the bolt head, which is 0.2 m. The output distance is the distance from the centre of the bolt head to where the force is applied, which is half of the bolt's diameter, or 3.0 mm. So, the I
 

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