The assumption that there is no energy loss is not accurate. Assuming the spring has a spring constant of 100Nm-1 and has a length change of 2 cm, the track is on a level surface, and that the final velocity of the car is 0.25 ms-1 when it reaches the end of the track, what is the percentage efficiency of the toy? Take the mass of the car to be 150 g.

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Answer: The answer is 23%, but I'm not sure how to get there.

Explanation: The assumption that no energy is lost is incorrect. What is the % efficiency of the toy if the spring has a spring constant of 100Nm-1 and a length change of 2 cm, the track is on a level surface, and the ultimate velocity of the car is 0.25 ms-1 when it reaches the end of the track? Assume that the automobile weighs 150 g.

The percentage efficiency of the toy will be 23%. It is the ratio of the kinetic and elastic potential energy,

What is elastic potential energy?

Elastic potential energy is the potential energy held when an elastic item is stretched or compressed by an external force, such as spring stretching.

It equals the work done to extend the spring, which is proportional to the spring constant k and the distance stretched.

Firstly, the kinetic energy is;

[tex]\rm KE = \frac{1}{2} mv^2 \\\\ \rm KE = \frac{1}{2} \times (150)(0.25)^2 \\\\ KE = 4.6875\ J[/tex]

The elastic potential energy is;

[tex]\rm E=0.5\times 4\times 10 = 20\\ E=100 \ J[/tex]

The percentage efficiency of the toy is found as;

[tex]\% \eta = \frac{4.6875 \times 100}{20} \\\\ \% \eta = 23 \% \\\\[/tex]

Hence, the percentage efficiency of the toy will be 23%

To learn more about the elastic potential energy, refer to the link;

https://brainly.com/question/12807194

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