I have a homework problem similar to 8.68 in Hugh Young\'sCollege Physics textbo

Question

I have a homework problem similar to 8.68 in Hugh Young'sCollege Physics textbook yet im unsure how to include coefficientsof friction between the bullet and block in my similar homeworkproblem. The problem is... Force of 200N is needed to compress the spring by 0.1meters. A spring positioned horizontally, with one of its endsattached to a wall. A 1 kg block is attached to the other endof the spring and is resting on the horizontal surface. A .01 kg bullet is fired horizontally towards the block and inthe collision the bullet is embedded into the block. Thecoefficient of friction between the block and horizontal surface is0.25. The maximal compression of the spring due to thecollision is .025 meters. Find the spring constant and initial velocity of thebullet. I found the spring constant to be 2000N/m by k=F/x     k= 200N/0.1m = 2000N/m I am unsure how to get initial velocity with friction alsoplaying a part in this problem. If i could get help thatwould be great. I have a homework problem similar to 8.68 in Hugh Young'sCollege Physics textbook yet im unsure how to include coefficientsof friction between the bullet and block in my similar homeworkproblem. The problem is... Force of 200N is needed to compress the spring by 0.1meters. A spring positioned horizontally, with one of its endsattached to a wall. A 1 kg block is attached to the other endof the spring and is resting on the horizontal surface. A .01 kg bullet is fired horizontally towards the block and inthe collision the bullet is embedded into the block. Thecoefficient of friction between the block and horizontal surface is0.25. The maximal compression of the spring due to thecollision is .025 meters. Find the spring constant and initial velocity of thebullet. I found the spring constant to be 2000N/m by k=F/x     k= 200N/0.1m = 2000N/m I am unsure how to get initial velocity with friction alsoplaying a part in this problem. If i could get help thatwould be great.

Explanation / Answer

Hooke's Law:
F = -kx k = -F/x = -(-200N)/0.1m = 2000N/m Conservation of Energy: Ei = Ef E = KE + U + ... Uspring = (1/2)kx2 Here, kinetic energy of the bullet is converted into potentialenergy of the spring plus work done by friction (neglect energyconverted into sound, heat, etc.) Ei = KEi + Ui =KEi + 0 = Ef = KEf + Uf= 0 + Uf + Wf KEi = Uf + Wf (1/2)mvi2 = (1/2)kx2 +Nx = (1/2)kx2 + (mg)x (1/2)mvi2 =(1/2)(2,000N/m)(0.025m)2 +(0.25)(1.01kg)(9.81m/s2)(0.025m) (1/2)mvi2 = 0.625 N*m + 0.062 N*m KEi = (1/2)mvi2 = 0.69J vi2 = 2(KE)/m = 2(0.69J)/(0.01kg) = 138(m/s)2 vi = 11.7 m/s vi = 11.7 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.