A 4.0-g bullet leaves the muzzle of a rifle with a speed of 340 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.75-m-long barrel of the rifle?
Given:
"m=0.004\\:\\rm kg"
"v_0=0,\\quad v=340\\:\\rm m\/s"
"d=0.75\\:\\rm m"
The energy-work theorem says
"\\frac{mv^2}{2}-\\frac{mv_0^2}{2}=Fd"Thus, the force exerted onj the bullet
"F=\\frac{mv^2}{2d}=\\frac{0.004*340^2}{2*0.75}=310\\:\\rm N"
Comments
Leave a comment