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I want to accelerate an object from resting state to a velocity $v_1$ then from a linear movement at speed $v_1$ to linear movement at speed $v_2$.

By conservation of energy law, the energy spent will be the difference between kinetic energy at the end and kinetic energy ar the beginning state. So for the first experiment I spend $\frac{1}{2}mv_1^2$ and for the second experiment $\frac{1}{2}m(v_2^2-v_1^2)$.

Now If I accelerate a resting object of mass 1Kg to a speed of $10ms^{-1}$ and then accelerate it to a speed of $20ms^{-1}$, the energy spent will be respectively 50J and 150J. Why does it take more energy to add $10ms^{-1}$ of velocity to a resting object than to an already moving object?

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1 Answers1

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Let's assume you are using the same constant force in each scenario. Then it takes the same amount of time to go from $0 \frac ms$ to $10 \frac ms$ as it does to go from $10 \frac ms$ to $20 \frac ms$ because the acceleration is constant. The work done in each case is $W=F\Delta x\space$. If your object is moving faster, then the object moves farther than a slower object in the same amount of time. This is why more work is done in the second case. You are applying the force across a longer distance.

This is also reflected in the power (rate at which work is done). $P=Fv$. Larger velocity for the same force gives you a larger rate of work. Of course, this power formula is just expressing what I talk about above, so this is not any new information.

BioPhysicist
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