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sertanlavr [38]
3 years ago
14

I just don't know how to do the question (a) and (b)​

Physics
1 answer:
ozzi3 years ago
5 0

Using the equation

F=ma

we can observe that you have to apply a non-zero net force to an object in order to make it accelerate. In fact, if the net force is zero you have

0=ma \iff a=0

Since we're assuming n\neq 0

Now, if the 12N force is applied, the object moves with a constant speed. A constant speed means no acceleration, since by definition the acceleration is a change in speed.

If this sounds counterintuitive to you (why I'm applying a force but I have to acceleration?) think of when we drive a car: even if you want to keep your speed constant, you still have to use the gas pedal, just enough so that the push of the motor balances exactly the road/wheels friction. If you give less gas, the friction becomes stronger, and the car slows down. If you give more gas, the motor push becomes stronger, and the car accelerates.

Back to your exercise: constant speed means to acceleration, so the net force must be zero. This implies that the friction force is exactly 12N.

If the force is increased to 18N, there will be a net force of 6N pushing the object, causing it to accelerate. Using again the same equation of before, and plugging the 3kg mass in the equation, we have

F=ma \iff 6=3a \iff a=2

So, the object moves with constant acceleration and initial speed of 10m/s for 0.2 seconds. It's final speed will be

v = v_0+at = 10+2\cdot 0.2 = 10.4

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Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in
USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

W=F\cdot d

Then, we can derive the magnitude of the force as:

F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N

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