How Much Power is Needed to Pull a Block Up an Incline?

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Homework Help Overview

The problem involves calculating the power required to pull a 1700 kg block of granite up an incline at a constant speed of 2.40 m/s, with a friction coefficient of 0.25 and an incline angle of 29°. The challenge lies in determining how to compute power without explicit time information.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between power, work, and force, questioning how to approach the calculation without a time variable. Some suggest using kinematics, while others propose alternative formulas for power that do not require time. There are inquiries about the correct application of formulas and the role of tension in the calculations.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided insights into the relationship between force, distance, and velocity, while others are still grappling with the implications of the given parameters. There is no explicit consensus yet, but productive dialogue is occurring around the concepts involved.

Contextual Notes

Participants note the absence of elapsed time as a source of confusion, emphasizing the need to rely on the provided velocity, mass, and friction coefficient to navigate the problem.

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A 1700 kg block of granite is pulled up an incline that has an angle of inclination θ = 29 ° with a constant speed of 2.40 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.25. How much power must be supplied by the winch?

How would you calculate power without a time? Does this involve using a lot of kinematics?
 
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chaotixmonjuish said:
A 1700 kg block of granite is pulled up an incline that has an angle of inclination θ = 29 ° with a constant speed of 2.40 m/s by a steam winch (see Figure). The coefficient of kinetic friction between the block and the incline is 0.25. How much power must be supplied by the winch?

How would you calculate power without a time? Does this involve using a lot of kinematics?

You could do this with kinematics...
but there is another formula for power (derived from the definition of work) that does not require an explicit time.
 
Oh, could you lead me to something that discusses the formula, just so I can get an intuitive understanding of the formula before utilizing it.
 
chaotixmonjuish said:
Oh, could you lead me to something that discusses the formula, just so I can get an intuitive understanding of the formula before utilizing it.

To get a feeling for the formula,
write (using \Delta x) the work done by a force F.
Now, write the power supplied by this force as a ratio in terms of this work.

To do this rigorously, one should use calculus.
 
I'm still having a few problems, but I would like to run some numbers by:

the tension in the rope pulling it up the incline would be 11719.7 N

and the rigorous formula, would that be:

m(1/2 Vf^2-1/2Vi^2)
 
Tension is right. But the formula is not...

You know the force of tension... so what is the work done by tension over a distance \Delta x

What is the work per unit time?
 
I would go about calculating time? Acceleration is zero.
 
chaotixmonjuish said:
I would go about calculating time? Acceleration is zero.

You don't need the time... what is \frac{\Delta x}{\Delta t} ?
 
We aren't given any elapsed time. This is where I'm confused. The only given is a velocity, mass, and coefficient of friction.
 
  • #10
chaotixmonjuish said:
We aren't given any elapsed time. This is where I'm confused. The only given is a velocity, mass, and coefficient of friction.

yup, you have velocity. use that.
 
  • #11
Could I break work into Force x Distance /time

but isn't distance/time a velocity

so could I use Force (the tension in the rope) times velocity
 
  • #12
chaotixmonjuish said:
Could I break work into Force x Distance /time

but isn't distance/time a velocity

so could I use Force (the tension in the rope) times velocity

exactly. the power is force*velocity.
 
  • #13
I guess my hints make more sense now.
More generally, it's
P=\vec F\cdot \vec v (assuming that these quantities remain constant during the motion).
 

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