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MAVERICK [17]
2 years ago
7

A student wearing frictionless in-line skates on a horizontal surface is pushed, starting from rest, by a friend with a constant

force of 60 N. How far must the student be pushed, so that her final kinetic energy is 300J?
Physics
2 answers:
grin007 [14]2 years ago
8 0

Answer:

Physics

Explanation:

Explanation:

We can use the Theorem of Work (W) and Kinetic Energy (K):

W=ΔK=Kf−Ki

it basically tells us that the work done on our system will show up as change in Kinetic Energy:

We know that the initial Kinetic Energy, Ki=12mv2i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get:

F⋅d=Ff

45d=352

and so:

d=35245=7.8≈8m

galben [10]2 years ago
3 0

Answer:

d= 352/45=7.8≈ 8m

Explanation

We can use the Theorem of Work (W) and Kinetic Energy (K):

W=ΔK=k <em>f</em> - k <em>i </em>, it is basically tells us that the work done on our system will show up as change in Kinetic Energy:We know that the initial Kinetic Energy, k<em>i</em> =1/2 <em>mv</em> 2/i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get: F . d = F<em>f</em>

45<em>d</em>= 352 and so d= 352/45=7.8 ≈ <em>8m</em>

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  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
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Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

  1. v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  3. The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>

<em></em>

<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>

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<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = <u><em>40.08 W</em></u>

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