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Yakvenalex [24]
4 years ago
7

A car with a mass of 1,324 kilograms, traveling at a speed of 20 meters/second, crashes into a wall and stops. What is the kinet

ic energy after this perfectly inelastic collision?
A. 0.0 joules
B. 3.2 × 103 joules
C. 1.0 × 104 joules
D. 1.3 × 104 joules ...?
Physics
2 answers:
Valentin [98]4 years ago
7 0
I think the correct answer from the choices listed above is option A. The kinetic energy after the perfectly inelastic collision would be zero Joules. <span>A </span>perfectly inelastic collision<span> occurs when the maximum amount of kinetic energy of a system is lost. Hope this answers the question.</span>
Ksenya-84 [330]4 years ago
5 0

Answer:

A. 0.0 joules

Explanation:

We can calculate the initial kinetic energy of the car, which is given by:

K=\frac{1}{2}mv^2 = \frac{1}{2}(1,324 kg)(20 m/s)^2=2.65\cdot 10^5 J

where m is the mass of the car and v its initial speed. After the collision, the car stops, so its final speed is zero. Therefore, the new kinetic energy of the car will be

K=\frac{1}{2}(1,324 kg)(0 m/s)^2 = 0 J

This type of collision is called 'perfectly inelastic collision', and it occurs when there is the maximum loss of kinetic energy, as in this case. Therefore, the correct answer is

A. 0.0 joules

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<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

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From the Newtons' equation of motion;

\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times  \frac{5280}{3600}  )^2 +2a\times 300 \\\\

The tangential accelerations are;

\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t =  -5.65 ft/sec^2 \\\\

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\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb

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The net or the total friction force exerted by the road on the tires at B. is found as;

\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

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