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### Re: Hovercraft B/C

Posted: March 15th, 2018, 2:39 pm
Do you need to know anything about initial and final velocities to solve this question?
Whoops: They are moving toward each other, the ball with a speed of 5 m/s and the pencil with a speed of 0.4 m/s.
Alright, first find the final velocities of both of the objects.

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

$v_{1i}-v_{2i}=v_{2f}-v_{1f}$

Plugging in the given values and solving for the final velocities gives 4.946 m/s for the ball and 10.346 m/s for the pencil. Now you can find the average force on the pencil:

$F\Delta t=m\Delta v=F*5=0.005(10.346-(-0.4))$

This gives an average force of 0.0107 N. You can check this answer by doing the same calculation on the ball (Newton's Third Law).
Yep, your turn! Wolfram alpha link to the series of equations if anyone wants it: [hardurl]http://www.wolframalpha.com/input/?i=(1 ... y+%2B+0.4)[/hardurl]

### Re: Hovercraft B/C

Posted: March 15th, 2018, 2:51 pm
Yep, your turn! Wolfram alpha link to the series of equations if anyone wants it: [hardurl]http://www.wolframalpha.com/input/?i=(1 ... y+%2B+0.4)[/hardurl]
Awesome

An amusement park is planning on constructing a new ride and needs help with all the planning. The track begins at a height of 60 meters and slopes downward at a constant angle. The coefficient of rolling friction for the wheels of the roller coaster is 0.05. If the park wants the ride to have a final velocity of at least 25 m/s at its lowest point, what should the minimum angle of slope be?

### Re: Hovercraft B/C

Posted: March 15th, 2018, 4:11 pm
An amusement park is planning on constructing a new ride and needs help with all the planning. The track begins at a height of 60 meters and slopes downward at a constant angle. The coefficient of rolling friction for the wheels of the roller coaster is 0.05. If the park wants the ride to have a final velocity of at least 25 m/s at its lowest point, what should the minimum angle of slope be?
$mgh = mg(60 m)$

$\frac12v^2 = mg(60 m)$

$v^2 = g(120 m)$

$\left(25 \frac{m}{s}\right)^2 = g(120 m) + 2ad$

$a = \frac{\left(25 \frac{m}{s}\right)^2 - g(120 m)}{2 * 60 m * csc \theta}$

$a_{friction} = 0.05a_{normal} = 0.05(g * cos \theta)$

$-0.05(g * cos \theta) = \frac{\left(25 \frac{m}{s}\right)^2 - g(120 m)}{2 * 60 m * csc \theta}$

$tan \theta = \frac{-0.05g * 120 m}{\left(25 \frac{m}{s}\right)^2 - g(120 m)}$

$\theta = \arctan \left(\frac{-0.05g * 120 m}{\left(25 \frac{m}{s}\right)^2 - g(120 m)}\right) = 0.106 = 6.09 \degree$

### Re: Hovercraft B/C

Posted: March 15th, 2018, 4:57 pm
An amusement park is planning on constructing a new ride and needs help with all the planning. The track begins at a height of 60 meters and slopes downward at a constant angle. The coefficient of rolling friction for the wheels of the roller coaster is 0.05. If the park wants the ride to have a final velocity of at least 25 m/s at its lowest point, what should the minimum angle of slope be?
$mgh = mg(60 m)$

$\frac12v^2 = mg(60 m)$

$v^2 = g(120 m)$

$\left(25 \frac{m}{s}\right)^2 = g(120 m) + 2ad$

$a = \frac{\left(25 \frac{m}{s}\right)^2 - g(120 m)}{2 * 60 m * csc \theta}$

$a_{friction} = 0.05a_{normal} = 0.05(g * cos \theta)$

$-0.05(g * cos \theta) = \frac{\left(25 \frac{m}{s}\right)^2 - g(120 m)}{2 * 60 m * csc \theta}$

$tan \theta = \frac{-0.05g * 120 m}{\left(25 \frac{m}{s}\right)^2 - g(120 m)}$

$\theta = \arctan \left(\frac{-0.05g * 120 m}{\left(25 \frac{m}{s}\right)^2 - g(120 m)}\right) = 0.106 = 6.09 \degree$
Correct! You're next!

### Re: Hovercraft B/C

Posted: March 15th, 2018, 5:24 pm
A 1 kg hovercraft is plugged into a power source with a 14 m long wire. It floats 2 cm off the ground and goes around in a circle around the power source at 6 m/s. The fans the hovercraft uses "convert" 1 Joule of energy into 1 Newton of force every second. Find the wattage of the power source.

### Re: Hovercraft B/C

Posted: March 16th, 2018, 12:06 am
A 1 kg hovercraft is plugged into a power source with a 14 m long wire. It floats 2 cm off the ground and goes around in a circle around the power source at 6 m/s. The fans the hovercraft uses "convert" 1 Joule of energy into 1 Newton of force every second. Find the wattage of the power source.
First find the power needed to keep the the hovercraft floating:

$P_1=\frac{mgh}{t}=\frac{1*9.8*0.02}{1}=0.196W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Adding the two gives an answer of [b]2.78 W[/b].

### Re: Hovercraft B/C

Posted: March 16th, 2018, 5:31 pm
A 1 kg hovercraft is plugged into a power source with a 14 m long wire. It floats 2 cm off the ground and goes around in a circle around the power source at 6 m/s. The fans the hovercraft uses "convert" 1 Joule of energy into 1 Newton of force every second. Find the wattage of the power source.
First find the power needed to keep the the hovercraft floating:

$P_1=\frac{mgh}{t}=\frac{1*9.8*0.02}{1}=0.196W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Adding the two gives an answer of [b]2.78 W[/b].
Does the lift fan really use more energy when the hovercraft is higher? Use a free-body diagram.

### Re: Hovercraft B/C

Posted: March 18th, 2018, 4:44 am
Does the lift fan really use more energy when the hovercraft is higher? Use a free-body diagram.
$P_1=\frac{mg}{t}=\frac{1*9.8}{1}=9.8W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Hope this is right   :oops:

### Re: Hovercraft B/C

Posted: March 18th, 2018, 8:18 am
Does the lift fan really use more energy when the hovercraft is higher? Use a free-body diagram.
$P_1=\frac{mg}{t}=\frac{1*9.8}{1}=9.8W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Hope this is right   :oops:
Do the hovercraft fans power the centripetal force or does the tension in the wire?

### Re: Hovercraft B/C

Posted: March 18th, 2018, 8:56 am
Does the lift fan really use more energy when the hovercraft is higher? Use a free-body diagram.
$P_1=\frac{mg}{t}=\frac{1*9.8}{1}=9.8W$

Next, find the centripetal force acting on the hovercraft:

$F_c=m\frac{v^2}{r}=1*\frac{6^2}{14}=2.57N$

$P_2=\frac{F_c}{t}=2.57W$

Hope this is right   :oops:
Do the hovercraft fans power the centripetal force or does the tension in the wire?
Is it just 9.8 W?