5748 Shares

Topic: Physics Help: Power Problem?**Question:**
I solved part A but cant solve part B. Thanks for help
A skier of mass 70.8 kg is pulled up a slope by a motor-driven cable.
(a) How much work is required for him to be pulled a distance of 59.2 m up a 30.4° slope (assumed frictionless) at a constant speed of 1.94 m/s?
This I found to be 20.785 kJ
b) A motor of what power is required to perform this task
Answer in hp

July 19, 2019 / By Rose

If it pulls him 59.2 meters at 1.94 meters per second, it takes 59.2 / 1.94 = 30.5 seconds 20.8 kjoules in 30.5 seconds is 0.681 kwatts 0r 681 watts. 1 hp = 746 watts, to this is 681/746 = 0.91 HP .

👍 120 | 👎 6

Did you like the answer? Work: force exerted over a distance: F x d Power: how much work per second since work is an energy we can say W = -U (where U is potential energy) U = mgh, at the top of the hill U = (56kg)(-9.8m/s2)(30.5m) = -16738.4 Joules so W = 16738.4 Joules now since you know how long it takes to get up to the top (10s) divide work by time to get power P = W/t = 16738.4 Joules/10 seconds = 1673.84 Watts or approximately 1.7 kiloWatts

Power is work divided by time. Your time is ( 59.2 m ) / ( 1.94 m/s ) = 30.5 seconds, so the power is ( 20.785 kJ ) / ( 30.5 sec ) = 681 Watts. Now convert that to horsepower: 1 hp = 746 Watts.

👍 40 | 👎 -1

Power = Work/Time You can find time from part (a) => Distance = speed*time => time = distance/speed = 59.2/1.94 = 30.52 s => Power = Work/time = (20.785*1000)/30.52.......(I multiplied work by 1000 to convert from kJ to J) => 681.089 W => now 1W = 0.001341 hp => Power = 681.089*0.001341 = 0.913335 hp

👍 35 | 👎 -8

Power is simply the total energy used divised by the time. T = time, X = distance moved, V=Velocity, E=total Energy P = Power = E.V / X

👍 30 | 👎 -15

The energy received by 1 m² = 1400 watt Energy required to be produced = 1 giga-watt = 1 x 10^12 The area reqd. = 1x10^12 / 1400 = 714285714 m² .................. ....................... ...........= 714 .28 Km² ................ ....................... ...............===========

If you have your own answer to the question Physics Help: Power Problem?, then you can write your own version, using the form below for an extended answer.
/**
* RECOMMENDED CONFIGURATION VARIABLES: EDIT AND UNCOMMENT THE SECTION BELOW TO INSERT DYNAMIC VALUES FROM YOUR PLATFORM OR CMS.
* LEARN WHY DEFINING THESE VARIABLES IS IMPORTANT: https://disqus.com/admin/universalcode/#configuration-variables*/
/*
var disqus_config = function () {
this.page.url = PAGE_URL; // Replace PAGE_URL with your page's canonical URL variable
this.page.identifier = PAGE_IDENTIFIER; // Replace PAGE_IDENTIFIER with your page's unique identifier variable
};
*/
(function() { // DON'T EDIT BELOW THIS LINE
var d = document, s = d.createElement('script');
s.src = 'https://help-study.disqus.com/embed.js';
s.setAttribute('data-timestamp', +new Date());
(d.head || d.body).appendChild(s);
})();
Please enable JavaScript to view the comments powered by Disqus.

Copyright 2019. All rights reserved | Read Questions and Answers 2019 Online