Power and Speed Relationship for Lifting Objects

In summary: P = mgh/t = 2*9.8*h/t = 78h/t = 78/19.6 = 4Note that h/t is the speed of the body.So you get a speed of 4 m/sec.So, in summary, the maximum power your body can deliver in lifting an object vertically is 78 W. To determine the speed at which you can lift a 2L full water container, you can use the equation P=\frac{m*g*h}{t} where P is power, m is mass, g is gravity, h is height (in this case, the height of the water container), and t is time. By substituting the given values, you
  • #1
woaini
58
0

Homework Statement



The maximum power your body can deliver in lifting an object vertically is 78 W. How fast could you lift, at constant speed, a 2L full water container?

P=78W
m=2kg
t=?
v=?


Homework Equations



P=ΔE/t
E=1/2mv2

The Attempt at a Solution



78W=(.5*m*v2)/t

t=(.5*2*v2)/78

I am unsure what to do when I have a speed constant.
 
Last edited:
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  • #2
Just remember that if you are lifting it at constant speed that means net force must be zero, or in other words your force should balance that of gravity. Then the rest is fairly straight forward calculus.

W is work done.

dW = F.dx
P = dW/dt
that implies P = F.(dx/dt)=F.v

Oh, and welcome to pf!
 
  • #3
I don't think I'm suppose to use calculus to solve this problem.

However if W is equal to what is done by gravity and the formula for P=W/t, this is what I get:

P=[itex]\frac{W}{t}[/itex] then t=[itex]\frac{m*g}{P}[/itex] which equals P=[itex]\frac{2*9.81}{78W}[/itex]=0.25s

Is this what you're trying to imply?
 
  • #4
woaini said:
I don't think I'm suppose to use calculus to solve this problem.

However if W is equal to what is done by gravity and the formula for P=W/t, this is what I get:

P=[itex]\frac{W}{t}[/itex] then t=[itex]\frac{m*g}{P}[/itex] which equals P=[itex]\frac{2*9.81}{78W}[/itex]=0.25s

Is this what you're trying to imply?

Don't confuse Work (Joules) with Power (Watts). Sometimes the variable W is used for Work, but there is also a unit called a Watt, denoted by W, which is not the same thing. A Watt is a unit of power, and represents a Joule/second. The given value, 78 W, means that you can produce 78 Watts of power for the described maneuver.

The current problem is asking you for "how fast" you can do something. In lay terms ("common language usage") that could mean in how short a time, but in physics "how fast" implies a speed. So you're looking for a speed. Look though your notes or text for a relationship between power and velocity... :wink:
 
  • #5
Ok so this is the relationship I found:

P = [itex]\frac{E}{T}[/itex] = [itex]\frac{0.5*m*v^2}{t}[/itex]

However, I do not have the t variable.

So this is essentially what I can narrow the equation down to:

78W = [itex]\frac{0.5*2*v^2}{t}[/itex] which equals to [itex]\frac{v^2}{t}[/itex]=78

I am unsure what to do when I am that this step as the seconds in the time cannot cancel out with the speed.
 
  • #6
Never mind, I think I understand it now.

With the equation P=[itex]\frac{E}{t}[/itex] the units are essentially [itex]\frac{kg*m^2}{s^3}[/itex]. This is equivalent to the product of F (N or [itex]\frac{kg*m}{s^2}[/itex]) and speed ([itex]\frac{m}{s}[/itex]). With this in my mind, the speed will be easy to determine as the force is known to be the product of mass and gravity.

Is there any flaw in this explanation?
 
  • #7
woaini said:
Never mind, I think I understand it now.

With the equation P=[itex]\frac{E}{t}[/itex] the units are essentially [itex]\frac{kg*m^2}{s^3}[/itex]. This is equivalent to the product of F (N or [itex]\frac{kg*m}{s^2}[/itex]) and speed ([itex]\frac{m}{s}[/itex]). With this in my mind, the speed will be easy to determine as the force is known to be the product of mass and gravity.

Is there any flaw in this explanation?

Nope, that is well reasoned. In fact, a very useful relationship is the fact that the power delivered by a force F to a body moving at speed V is P = F*V :smile:
 
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  • #8
woaini said:
I don't think I'm suppose to use calculus to solve this problem.

However if W is equal to what is done by gravity and the formula for P=W/t, this is what I get:

P=[itex]\frac{W}{t}[/itex] then t=[itex]\frac{m*g}{P}[/itex] which equals P=[itex]\frac{2*9.81}{78W}[/itex]=0.25s

Is this what you're trying to imply?

yes, but the work done by gravity is m*g*h. so your expression becomes

P=[itex]\frac{m*g*h}{t}[/itex]

Here mg is the weight and h/t is the change in height with time, or simply your required velocity

woaini said:
Ok so this is the relationship I found:

P = [itex]\frac{E}{T}[/itex] = [itex]\frac{0.5*m*v^2}{t}[/itex]

However, I do not have the t variable.

So this is essentially what I can narrow the equation down to:

78W = [itex]\frac{0.5*2*v^2}{t}[/itex] which equals to [itex]\frac{v^2}{t}[/itex]=78

I am unsure what to do when I am that this step as the seconds in the time cannot cancel out with the speed.

The only energy change is gravitational potential, since Kinetic Energy is constant due to velocity being constant.

W = ΔE = m*g*h

woaini said:
Never mind, I think I understand it now.

With the equation P=[itex]\frac{E}{t}[/itex] the units are essentially [itex]\frac{kg*m^2}{s^3}[/itex]. This is equivalent to the product of F (N or [itex]\frac{kg*m}{s^2}[/itex]) and speed ([itex]\frac{m}{s}[/itex]). With this in my mind, the speed will be easy to determine as the force is known to be the product of mass and gravity.

Is there any flaw in this explanation?

no, not at all but it could lead to more confusion in other situations. Like gneill said and I showed it above

a very useful relationship is the fact that the power delivered by a force F to a body moving at speed V is P = F*V
 

What is work problem in watts?

The work problem in watts is a physics concept that refers to the amount of energy used by an object in a certain amount of time. It is measured in watts, which is a unit of power equal to one joule per second.

How is work problem in watts calculated?

The work problem in watts is calculated by multiplying the force applied to an object by the distance it moves in the direction of that force, and then dividing by the time it takes to complete the work. This can be represented by the equation W = F*d/t.

What are some real-life examples of work problem in watts?

Some real-life examples of work problem in watts include pushing a cart up a hill, lifting a heavy object, and riding a bicycle. In each of these scenarios, a certain amount of force is applied over a distance, resulting in work being done over a specific amount of time.

What is the relationship between work problem in watts and power?

Work problem in watts is a measure of power, which is the rate at which work is done. The higher the amount of work done in a given amount of time, the greater the power. Therefore, the higher the watts, the more powerful the object or system is.

How is work problem in watts useful in the scientific field?

Work problem in watts is a fundamental concept in physics and is used to understand and analyze the behavior of objects and systems in the physical world. It is also used in various engineering and technology applications, such as designing efficient machines and calculating energy usage.

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