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## anonymous one year ago a

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1. anonymous

@Michele_Laino

2. Michele_Laino

is the initial speed of the bullet: 3*104=312 m/sec?

3. Michele_Laino

oops..312 cm/sec?

4. Michele_Laino

Here, the work done by the friction force, has to be equal to the kinetic energy of the bullet, so we can write this: $\Large F \times d = \frac{1}{2}mv_0^2$ where F is the requested force, d= 0.05 meters, and m=0.002 Kg, v_0=3.12 m/sec

5. Michele_Laino

so, dividing both sides by d, we get: $\Large F = \frac{{mv_0^2}}{{2d}} = \frac{{0.002 \times {{3.12}^2}}}{{2 \times 0.05}} = ...Newtons$

6. Michele_Laino

that's right!

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