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(D) According to the work-energy theorem,the work done (energy required) is equal to the change in kinetic energy.For the first case,the energy requir

Vedclass · Accepted Answer

$3$ times

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(D) According to the work-energy theorem,the work done (energy required) is equal to the change in kinetic energy.For the first case,the energy required to accelerate from $v_1 = 10 \ m/s$ to $v_2 = 20 \ m/s$ is:$E_1 = \frac{1}{2} m (v_2^2 - v_1^2) = \frac{1}{2} m (20^2 - 10^2) = \frac{1}{2} m (400 - 100) = \frac{1}{2} m (300)$.For the second case,the energy required to accelerate from rest $(v_0 = 0 \ m/s)$ to $v_1 = 10 \ m/s$ is:$E_2 = \frac{1}{2} m (v_1^2 - v_0^2) = \frac{1}{2} m (10^2 - 0^2) = \frac{1}{2} m (100)$.Taking the ratio:$\frac{E_1}{E_2} = \frac{\frac{1}{2} m (300)}{\frac{1}{2} m (100)} = 3$.Therefore,the energy required is $3$ times.

The energy required to accelerate a car from $10 \ m/s$ to $20 \ m/s$ is how many times the energy required to accelerate the same car from rest to $10 \ m/s$?

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