The driving force exerted by D's engine is 4 times as great when D is at the top of the hill as it is when D is at the bottom. Find the ratio of the power developed by D's engine at the top of the hill to the power developed at the bottom.
Thanks in advance!!
You're more likely to get more accurate help if you post this question on the physics help forum, our sister site. I say that for your benefit.
Since power is defined as in work per unit time, and work is force times distance, then power = force time velocity. You tell us the engine develps 4 times as much force at the top of the hill as at the bottom, but you don't tell us about the velocity - specifically whether the velocity of D is the same at the top of the hill as at the botttom. Without that information it's impossible to answer.
Opps, sorry, I forgot the previous parts
A car C of mass 1200kg climbs a hill of length 500m at a constant speed. The hill is inclined at an angle of 6 degrees to the horizontal. The driving force exerted by C's engine has magnitude 1800N. Another car D, also of mass 1200kg, climbs the same hill with increasing speed. The speed at the bottom is 8m/s and the speed at the top is 20m/s. It is assumed that the resistance to the motion of D is constant and has magnitude 700N. The driving force exerted by D's engine is 4 times as great when D is at the top of the hill as it is when D is at the bottom. Find the ratio of the power developed by D's engine at the top of the hill to the power developed at the bottom.
Thanks in advance!!
Find what ratio of power? Thrust to weight ratio?Originally Posted by Cathelyn13
Opps, sorry, I forgot the previous parts
A car C of mass 1200kg climbs a hill of length 500m at a constant speed. The hill is inclined at an angle of 6 degrees to the horizontal. The driving force exerted by C's engine has magnitude 1800N. Another car D, also of mass 1200kg, climbs the same hill with increasing speed. The speed at the bottom is 8m/s and the speed at the top is 20m/s. It is assumed that the resistance to the motion of D is constant and has magnitude 700N. The driving force exerted by D's engine is 4 times as great when D is at the top of the hill as it is when D is at the bottom. Find the ratio of the power developed by D's engine at the top of the hill to the power developed at the bottom.
Thanks in advance!!
Net work done, W by the engine = 1800x500 = 9x10^5 J
W =change in potential energy of car, W1 + work done against resistance, W2
W1 = 1200x9.8x500x sin 6 = 6.1463x10^5 J
W2 = 2.8537x10^5 J is the answer to th first problem.
In the second problem, we have to find out the net external work done, WE, by the engine.. It has three components. Change in kinetic energy of car, KE, change in potential energy, PE and work done against resistance.
WE = 0.5x1200(20^2-8^2) + 6.1463x10^5 + 700x500
WE = 2.016x10^5+6.1463x10^5+3.5x10^5 = 1.1623x10^6 J is the second answer.
lol, its been 2 years since you asked this question :P anyways..
A car C of mass 1200kg climbs a hill of length 500m at a constant speed. The hill is inclined at an angle of 6 degrees to the horizontal.
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HOLY CRAP!! I just figured that the Physics forum fell under the "mash hammer" and disappeared.
Gee, I'm a mod on that forum too. Maybe I ought to visit it once in a while.
Thanks for the info!
-Dan
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