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3 years ago

A 13561 N car traveling at 51.1 km/h rounds

Physics

1 answer:

Minchanka [31]3 years ago

3 0

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

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A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal.

Elza [17]

Answer:

Normal force: 167.48 N

Explanation:

First of all it is necessary to draw the free body diagram of the suitcase adding alll the forces stated on the question: the normal force, the friction force and pull force exterted by the woman. Additionally, we need to add the weight, the forces exerted by Earth's gravity. I attached the diagram so you can check it.
We need to resolve all the unknown quantities on this exercise, so we need to write down the sum of forces equations on X-Axis and Y-axis. Remember that force exerted by the woman has an angle with respect the horizontal (X-Axis), that is to say it has force compoents on both X and Y axis. The equations will be equal to zero since the suitcase is at constant speed (acceleration is zero).

∑ $F_{x}$ : $F{x}-20 = 0$

∑ $F_{y}$ : $N -W+F_{y}=0$

Our objetive is to find the value of the normal force. It means we can solve the sum of Y-axis for N. The solution would be following:

$N = W - Fy$

Keep in mind Weight of the suitcase (W) is equal to the suitcase mass times the acceleration caused by gravity (9.81 $\frac{m}{s^{2}}$ . Furthermore, Fy can be replaced using trigonometry as $Fsin(\theta)$ where θ is the angle above the horizontal. So the formula can be written in this way:

$N = mg -Fsin(\theta)$

We need to find the value of θ so we can find the value of N. We can find it out solving the sum of forces on X-axis replacing <em>Fx</em> for Fcos(θ). The equation will be like this:

$Fcos(\theta) -20 = 0$ ⇒ $Fcos(\theta) = 20$ $\theta [tex]\theta=cos^{-1}(20/F)$

Replacing the value of F we will see θ has a value of 55.15°. Now we can use this angle to find the value of N. Replacing mass, the gravity acceleration and the angle by their respective values, we will have the following:

$N = 20 x 9.81 - 35sin(55.15)$ ⇒ $N = 167.48 N$

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Explanation:

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The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.

In short, Your Answer would be "Decreases"

Hope this helps!</span>

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