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Varvara68 [4.7K]
3 years ago
8

A car, X, with a mass of 1500 kg is moving west with an acceleration of 0.50 m/s2. A second car, Y, of equal mass is heading eas

t with an acceleration of 1.0 m/s2. Which of the following statements best describes the net force acting on each car?
Physics
1 answer:
ehidna [41]3 years ago
8 0

Answer:

A

Explanation:

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Answer:

<em>The required constant friction force for the last 20 m is 6,862.8 N</em>

Explanation:

<u>Energy Conversion</u>

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

E_1=K_1+W_1

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J

This means that the kinetic energy that remains when the car reaches ground level is

K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

W_2=F_{r2}\cdot x_2=137,256

Solving for Fr2

\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N

The required constant friction force for the last 20 m is 6,862.8 N

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<span>F = m*a = 7000kg * 9.8N/kg = 68,600 N. 

68,600 N is your answer
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