A 12 g bullet is fired into a 9.0 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for wood sliding on wood is 0.20. What was the speed of the bullet? (please help, I've tried setting this problem up multiple times and can't get it right)

Hey! We 've verified this expert answer for you, click below to unlock the details :)

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

|dw:1352789469767:dw||dw:1352789574359:dw| |dw:1352789707768:dw||dw:1352789857307:dw| Using \( v^2=u^2+2as\) \({v_4}^2={v_3}^2-2as\) Here |dw:1352790096724:dw| Then using conservation of momentum principle \(mv+m_1v_1=(m+m_1)v_3\) \(0.012v=9.012*2as\) |dw:1352790355574:dw|

you might have also used work energy theorem simply : W(friction) = 1/2 (m1 v^2) where m1 = mass of bullet => U (m1+m2) g . d = 1/2 m1 v^2 m2 = mass of block U=coefficient of friction d= distance substituting, we have v^2 = (2 * 0.2 * 9.012 *10 * 0.05)/(0.012) please correct me if am wrong somewhere as i am not too strong with this topic..

in this topic*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.