A rifle fires a 20 gram bullet, at a speed of 600m/s into a wooden block of 4kg

Question

A rifle fires a 20 gram bullet, at a speed of 600m/s into a wooden block of 4kg resting at the bottom of a 30 degree incline. Assuming no friction, the bullet/block moves up the incline approximately 182.75m before coming, instantaneously, to rest.

Once it reaches 182.75m, the bullet explodes and breaks the block into two equal parts. One part is driven up the incline; the other down the incline (assuming that the bullet has disintegrated and the massis ignored). Both pieces have the same speed of 7.0 m/s.

How war from 182.75m will the upwardly moving piece go if that part of the incline has a coefficient of friction of 0.4?

Explanation / Answer

Given that

The initial speed of the piece vi = 7.0 m/s

and coefficent of friction = 0.4

Then from work energy theorem

       W = K

   - mg cos = 0 - 1/2 mv2

       [(0.74)(2kg)(9.8 m/s2) cos30] l = (0.5)(2 kg) (7 m/s)2

Therefore the distance

                    l = 24.5 / 6.28

                      = 3.9 m

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