What is the Average Power of a Sled Being Pulled with a Constant Force?

In summary: I think it would be good if everything was in watts but I'm not sure. Do you have any thoughts on that?Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:In summary, the homework statement is that a sled is being pulled along a horizontal surface by a horizontal force. During the experiment, a displacement was found. The force is 600 N and the power is P = Fs/
  • #1
ssb
119
0

Homework Statement



A sled is being pulled along a horizontal surface by a horizontal force F of magnitude 600 N. Starting from rest, the sled speeds up with acceleration 0.08 m/s^2 for 1 minute.

Find the average power P created by force F.


Homework Equations



P = Fs/T

The Attempt at a Solution



So I need to first find displacement of movement. If y = .08x^2, then at 60 seconds y = 288 ... so displacement is 288 right?

The force is 600 N
so since P = Fs/T, then (600*288)/60 should be my answer right? Also would this answer be in watts? I know that 1 joule/second = 1 watt.
 
Physics news on Phys.org
  • #2
Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:
 
  • #3
neutrino said:
Apart from a tiny factor of 1/2, everything else is right. Yes, the SI unit of power is the Watt. Can you find where you missed the factor? :wink:

So y = (1/2).08x^2

this would yield a displacement of 144
so then (600*144)/60

would be a better answer?

Also this answer is in joules so I need to multiply the whole thing by 60 to yield watts right?

so my final answer in watts should be 600*144 ?
 
  • #4
ssb said:
So y = (1/2).08x^2

this would yield a displacement of 144
so then (600*144)/60

would be a better answer?
That is the correct answer.

Also this answer is in joules so I need to multiply the whole thing by 60 to yield watts right?

As I said, there was nothing wrong with your first answer apart from that 0.5. It is Newtons.metre/second -> Joule/second -> Watt. Moreover, why would you take the trouble of first dividing and then immediately multiplying by 60?
 
  • #5
neutrino said:
That is the correct answer.



As I said, there was nothing wrong with your first answer apart from that 0.5. It is Newtons.metre/second -> Joule/second -> Watt. Moreover, why would you take the trouble of first dividing and then immediately multiplying by 60?

ack! :blushing: :blushing: :redface:

Thanks buddy! I appreciate the help!
 
  • #6
is power factor good or bad? Why?

Im just not sure!
 

1. What is the average power of a sled?

The average power of a sled is the amount of work done per unit time, or the rate at which energy is expended, while pulling or pushing the sled. It is typically measured in watts (W).

2. How is the average power of a sled calculated?

The average power of a sled can be calculated by dividing the total work done (in joules) by the time taken to complete the task (in seconds). This can also be expressed as the product of force (in newtons) and velocity (in meters per second).

3. What factors affect the average power of a sled?

The average power of a sled is affected by various factors including the mass of the sled, the force applied to move it, and the speed at which it is pulled. The surface on which the sled is being pulled also plays a role, as does the efficiency of the sled's design.

4. How does the average power of a sled impact performance?

The average power of a sled can impact performance in various ways. A higher average power generally leads to faster speeds and greater distances covered in a given amount of time. It also affects the efficiency and endurance of the sled and its rider.

5. Can the average power of a sled be improved?

Yes, the average power of a sled can be improved through various methods such as optimizing the sled's design, using stronger and more efficient materials, and improving the technique and strength of the rider. Technology, such as adding gears or motors, can also enhance the average power of a sled.

Similar threads

  • Introductory Physics Homework Help
Replies
29
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
4
Views
4K
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
4K
Back
Top