Power & Velocity Homework Solution | 450 kW

In summary: Thank you for the fast reply.Okay. I was suspecting much. But how do I do the integration for the ## \frac{1}{v} ## with respect to ds? Any clues as to where to start?Use [itex]F= ma= m\frac{dv}{dt}= m\frac{dv}{ds}\frac{ds}{dt}= mv\frac{dv}{ds}[/itex]
  • #1
elitewarr
63
0

Homework Statement


The rocket sled has a mass of 4 Mg and travels along the smooth horizontal track such that it maintains a constant power output of 450 kW. Neglect the loss of fuel mass and air resistance, and determine how far the sled must travel to reach a speed of v = 60 m>s starting from rest.

Homework Equations



P = Fv
## E_k = \frac{1}{2} mv^2 ##
W=Fs

The Attempt at a Solution



Since the rocket sled is moving along a flat surface, and that there is no loss of energy, we can conclude that the energy coming from the power output will be converted to kinetic energy.

## E_k = W = Fs = \frac{P}{v} s ##

Thus simplifying, we get

## \frac{1}{2} mv^3 = Ps ##

But apparently, there is something wrong with the equations since if we go about from another way,

## P = Fv = (ma)v = (mv\frac{dv}{ds})v ##

## \int_0^s P \, ds = m\int_0^v v^2 \, dv ##

## Ps = \frac{1}{3} mv^3 ##

Is there something wrong with my concept or that I am not seeing?

Thank you.
 
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  • #2
elitewarr said:
W=Fs
Force will not be constant here, so that needs to be an integral, not a simple product.
 
  • #3
haruspex said:
Force will not be constant here, so that needs to be an integral, not a simple product.

Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the ## \frac{1}{v} ## with respect to ds? Any clues as to where to start?
 
Last edited by a moderator:
  • #4
might be useful: Energy = power x time
 
  • #5
elitewarr said:
Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the ## \frac{1}{v} ## with respect to ds? Any clues as to where to start?
Since v= ds/dt, [itex]\int \frac{P}{v}ds= P\int \d
elitewarr said:
Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the ## \frac{1}{v} ## with respect to ds? Any clues as to where to start?
Since [itex]v= \frac{ds}{dt}[/itex],
[tex]P\int \frac{1}{v}ds= P\int \frac{1}{\frac{ds}{dt}}ds= P\int \frac{dt}{ds}ds= P\int dt= P(t_1- t_0)[/tex]
 
  • #6
elitewarr said:
Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the ## \frac{1}{v} ## with respect to ds? Any clues as to where to start?
Relate F to dv/dt.
Edit: or better still, to vdv/ds.
 
Last edited:

1. What is power?

Power is the rate at which energy is transferred or converted from one form to another. In simpler terms, it is the amount of work done per unit of time.

2. How is power measured?

Power is measured in watts (W), which is a unit of energy per unit of time. A larger unit of power is kilowatt (kW), which is equal to 1000 watts.

3. What is the equation for power?

The equation for power is: power = work/time. In other words, power is equal to the amount of work done divided by the time it took to do that work.

4. What is the relationship between power and velocity?

Power and velocity are directly proportional. This means that as velocity increases, power also increases. This is because an increase in velocity requires more work to be done in the same amount of time.

5. How do you solve for power with a given velocity?

To solve for power with a given velocity, you can use the equation: power = force x velocity. This equation relates power to force and velocity, and can be used to calculate power by multiplying the force applied with the velocity at which it is applied.

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