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Bond [772]
2 years ago
11

A car turns into a driveway that slopes upward at a 9 degree angle, car is moving at 6.5 m/s. If the driver lets the car coast,

how far along the slope will the car roll before being instantaneously at rest and start rolling back? Please show kinematic steps. Thanks!
Physics
2 answers:
expeople1 [14]2 years ago
6 0

To put it in the simplest form, the automatic transmission has a torque converter that uses the transmission fluid to turn the other side of the torque converter. Ex. Putting two room fans facing each other and turning one on and wind flowing from one fan makes the second fan facing the rotating fan turn and the idling engine probably doesn't have enough power to hold it.

I guess more technically, in an automatic, the car selects a neutral gear (no gear) when it is at rest, and the brake pedal is pressed. Upon releasing the brake, the car will apply a small forward force that will hold the car steady on a very small gradient or propel it forwards slowly on a flat or downhill gradient. This force is only started when the brake is being released, as this is the indicator for the car to change into 1st gear.

In a situation where a car is rolling back down the hill it is facing up, there could be one or more of several situations at hand.

Most likely, it simply lacks power to hold on the gradient of the slope. In this case, you'll have to perform a hill start to maintain brake force until forward propulsion is enough to move the car forward. Hill starts are almost always necessary for manual transmission cars. Additionally, it could be the driver's pedal transfer from brake to accelerator is to slow. That would also account for some part of the slip.

Okay, what's missing here is the clutch. The clutch as I'm sure you're aware is a series of plates that connect to each other and transfer power due to friction. The amount of friction is adjustable depending on how much force is applied to hold the plates together (or apart). It is easier to see this in a manual car at low speeds, where the driver hovers the clutch actuation pedal around the "Friction Point". This point is the fine line between moving (increased friction between clutch plates holds them together more firmly, thus transferring more power) and staying stationary (clutch plates disengaged from each other). From the Friction Point, any further release of the clutch will cause the car to move forwards because the transmission is engaging with more of the engine's power. Depressing the clutch pedal back in will not have any effect, as it will just keep the clutch plates separate.

In an automatic car, this is all controlled by computer algorithms, determining how much the clutch should be engaged to reach a certain speed. Taking off from the lights on a hill for example will not necessarily register as any different to taking off on flat ground. The effect of this is that the car is assuming that is requires a certain number of revs and a predetermined clutch setting to accelerate smoothly. Due to the increased force the hill provides, the car will move backwards until the power again reaches a level that will overcome its slippage.

One other thing, is that clutches work both ways. Since only friction holds them together, the torque exerted by the wheels back through the drive-train to the clutch can cause the plates to slip when they are not completely engaged. This results in the wheels moving independently of the running engine, as the clutch is separating the forces they would exert on each other.

I think the key thing to note is that despite not having a pedal to operate it, Automatic cars still have a clutch - just one that relies on a computer to function.

If it's not the engine or the clutch which are both behaving as normal, and we're still assuming it's an auto, there could be a problem with the car's computer system, transmission, clutch or gearbox that is causing the slippage. I would assume this is less likely, but it might be worth checking if it happens to your car. your frickin' welcome

Solnce55 [7]2 years ago
4 0
<span>Agreed, so you need to draw a diagram to start. That is the car on the driveway that has a nine degree slope. From there , you can then draw a free body diagram to analyze the forces on the car. Give it a try please.

You have the ability to answer the problem by your own.

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!

</span>
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A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

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3 years ago
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CaHeK987 [17]
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You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

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The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

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To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

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PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

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\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

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