How much work can be realized? Simple physics I can't understand.

In summary, the hammer of a machine needs to have a mass of 1.00 x 10² kg in order to break cement. The machine's motor lifts the hammer to a height of 5.00 meters every 10.0 seconds. To find the work done in 1.00 minute, we use the equation W = Fd and convert from seconds to minutes, resulting in 29430 J of work done by the machine in 1.00 minute.
  • #1
fixedglare
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Homework Statement


To be able to break cement, the hammer of a machine needs to have a mass of 1.00 x 10² kg.
The motor of the machine lifts it at a height of 5.00 meters every 10.0 seconds.

a) How much work did the machine realize within 1.00 minute?


Homework Equations



W = Fd

1 min = 60s/1min

The Attempt at a Solution



Work = 100 kg * 5.00 m = 500 J

60 seconds * 500 J = 30000 J

Is that how its done? I'm pretty confused.
 
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  • #2
You have the right idea. The first part:
fixedglare said:
Work = 100 kg * 5.00 m = 500 J
is the work done in lifting the block 5.00m, and the problem tells it lifts the hammer 5.00m in 10.0s, so you can also say that's the work done in 10.0s. Then, since the answer needs to be the work done over a full minute, you just have to make the conversion from work done in 10.0s to work done in one minute.
 
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  • #3
jackarms said:
You have the right idea. The first part: is the work done in lifting the block 5.00m, and the problem tells it lifts the hammer 5.00m in 10.0s, so you can also say that's the work done in 10.0s. Then, since the answer needs to be the work done over a full minute, you just have to make the conversion from work done in 10.0s to work done in one minute.

So, 500 J * 60 seconds = 30000 s

then

30000 s * 1 min /60 s?
 
  • #4
'Couple of comments:

1.00 × 102 kg is the hammer's mass, not its weight. Weight is the measure of force. :smile:

The hammer is raised once every 10 seconds (not every second, but once every 10 seconds). So how many times is the hammer raised in a minute?
 
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  • #5
No, you're mixing up your units. What you have is this:$$500J \cdot \frac{1}{10s}$$That's the work done per second, or the rate at which work is done. Then you have find the work done in a full minute. That means using this relationship:$$ amount = rate \cdot time$$
Or, just make a proportion:$$\frac{500J}{10s} = \frac{?}{1min}$$, where [itex]?[/itex] is the amount of work you're looking for.

EDIT: Yes, what collinsmark said. That 500J needs to be recalculated using weight instead of mass.
 
  • #6
collinsmark said:
'Couple of comments:

1.00 × 102 kg is the hammer's mass, not its weight. Weight is the measure of force. :smile:

The hammer is raised once every 10 seconds (not every second, but once every 10 seconds). So how many times is the hammer raised in a minute?

Weight = 1.00 x 10 2 kg * 9.81 m/s = 981 N

Work = 981 N * 5.00 m = 4905 J

in 1 minute the hammer realizes 6 * 4905 J = 29430 J?
 
  • #7
fixedglare said:
Weight = 1.00 x 10 2 kg * 9.81 m/s = 981 N

Work = 981 N * 5.00 m = 4905 J

in 1 minute the hammer realizes 6 * 4905 J = 29430 J?

There you go. :approve: That looks right to me. :smile:
 
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What is work in physics?

Work in physics is defined as the amount of energy transferred by a force acting through a distance. It is a scalar quantity and is measured in joules (J).

How is work calculated?

The formula for calculating work is W = Fd cosθ, where W is work, F is the force applied, d is the distance moved, and θ is the angle between the force and the displacement.

What is the unit of work?

The unit of work is joule (J), which is also equivalent to a newton-meter (N*m).

What factors affect the amount of work that can be realized?

The amount of work that can be realized is affected by the magnitude of the force, the distance over which the force is exerted, and the angle between the force and the displacement.

Can work be negative?

Yes, work can be negative if the force and displacement are in opposite directions. This means that the force is working against the motion and is taking energy away from the object.

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